When Should I Use Taylor Series for Limits?












1












$begingroup$


I get confused between when to apply L'Hospital Rule and Taylor Series.

Is there any set of trigger points in the questions, that would be easier to solve with Taylor Series?



For Example, If the denominator is in terms of a large power of $x (>3)$, then L'Hospital Rule usually becomes complicated and is not advised.



Edit:
Solve



$lim _{xto 0}left(left(sin xright)^{frac{1}{x}}+left(frac{1}{x}right)^{sin x}right)$



The options given are: $0, 1, -1, infty$






calculus limits taylor-expansion limits-without-lhopital







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you have a specific example? You're generalising too much when there are a lot of limits you could simplify with L'Hopital's or Taylor series.
    $endgroup$
    – Toby Mak
    Dec 9 '18 at 5:29












  • $begingroup$
    @TobyMak, I meant the limits where L'Hospital Rule cannot be applied. I'll add an example in a few minutes. I just wanted to know if there are certain situations where we should use Taylor Expansions.
    $endgroup$
    – Harshil Bhatt
    Dec 9 '18 at 5:34












  • $begingroup$
    My point still holds. Generally speaking, if there are trigonometric functions such as $sin x, cos x, ln x$, or exponential functions such as $e^x$ and $ln x$, then the Taylor expansions will be easier.
    $endgroup$
    – Toby Mak
    Dec 9 '18 at 5:36












  • $begingroup$
    Have a look to this small nightmare : math.stackexchange.com/questions/925916/…
    $endgroup$
    – Claude Leibovici
    Dec 9 '18 at 6:30










  • $begingroup$
    You should first try manipulating the given expression to make use of standard limits. Most limit problems on this website don't need advanced tools like L'Hospital's Rule or Taylor. Even when these tools are needed the prior manipulation greatly simplifies the use of these tools.
    $endgroup$
    – Paramanand Singh
    Dec 9 '18 at 14:01
















1












$begingroup$


I get confused between when to apply L'Hospital Rule and Taylor Series.

Is there any set of trigger points in the questions, that would be easier to solve with Taylor Series?



For Example, If the denominator is in terms of a large power of $x (>3)$, then L'Hospital Rule usually becomes complicated and is not advised.



Edit:
Solve



$lim _{xto 0}left(left(sin xright)^{frac{1}{x}}+left(frac{1}{x}right)^{sin x}right)$



The options given are: $0, 1, -1, infty$






calculus limits taylor-expansion limits-without-lhopital







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you have a specific example? You're generalising too much when there are a lot of limits you could simplify with L'Hopital's or Taylor series.
    $endgroup$
    – Toby Mak
    Dec 9 '18 at 5:29












  • $begingroup$
    @TobyMak, I meant the limits where L'Hospital Rule cannot be applied. I'll add an example in a few minutes. I just wanted to know if there are certain situations where we should use Taylor Expansions.
    $endgroup$
    – Harshil Bhatt
    Dec 9 '18 at 5:34












  • $begingroup$
    My point still holds. Generally speaking, if there are trigonometric functions such as $sin x, cos x, ln x$, or exponential functions such as $e^x$ and $ln x$, then the Taylor expansions will be easier.
    $endgroup$
    – Toby Mak
    Dec 9 '18 at 5:36












  • $begingroup$
    Have a look to this small nightmare : math.stackexchange.com/questions/925916/…
    $endgroup$
    – Claude Leibovici
    Dec 9 '18 at 6:30










  • $begingroup$
    You should first try manipulating the given expression to make use of standard limits. Most limit problems on this website don't need advanced tools like L'Hospital's Rule or Taylor. Even when these tools are needed the prior manipulation greatly simplifies the use of these tools.
    $endgroup$
    – Paramanand Singh
    Dec 9 '18 at 14:01














1












1








1





$begingroup$


I get confused between when to apply L'Hospital Rule and Taylor Series.

Is there any set of trigger points in the questions, that would be easier to solve with Taylor Series?



For Example, If the denominator is in terms of a large power of $x (>3)$, then L'Hospital Rule usually becomes complicated and is not advised.



Edit:
Solve



$lim _{xto 0}left(left(sin xright)^{frac{1}{x}}+left(frac{1}{x}right)^{sin x}right)$



The options given are: $0, 1, -1, infty$






calculus limits taylor-expansion limits-without-lhopital







share|cite|improve this question











$endgroup$




I get confused between when to apply L'Hospital Rule and Taylor Series.

Is there any set of trigger points in the questions, that would be easier to solve with Taylor Series?



For Example, If the denominator is in terms of a large power of $x (>3)$, then L'Hospital Rule usually becomes complicated and is not advised.



Edit:
Solve



$lim _{xto 0}left(left(sin xright)^{frac{1}{x}}+left(frac{1}{x}right)^{sin x}right)$



The options given are: $0, 1, -1, infty$






calculus limits taylor-expansion limits-without-lhopital




calculus limits taylor-expansion limits-without-lhopital






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 17:43









gimusi

92.8k84494




92.8k84494










asked Dec 9 '18 at 5:06









Harshil BhattHarshil Bhatt

94




94












  • $begingroup$
    Do you have a specific example? You're generalising too much when there are a lot of limits you could simplify with L'Hopital's or Taylor series.
    $endgroup$
    – Toby Mak
    Dec 9 '18 at 5:29












  • $begingroup$
    @TobyMak, I meant the limits where L'Hospital Rule cannot be applied. I'll add an example in a few minutes. I just wanted to know if there are certain situations where we should use Taylor Expansions.
    $endgroup$
    – Harshil Bhatt
    Dec 9 '18 at 5:34












  • $begingroup$
    My point still holds. Generally speaking, if there are trigonometric functions such as $sin x, cos x, ln x$, or exponential functions such as $e^x$ and $ln x$, then the Taylor expansions will be easier.
    $endgroup$
    – Toby Mak
    Dec 9 '18 at 5:36












  • $begingroup$
    Have a look to this small nightmare : math.stackexchange.com/questions/925916/…
    $endgroup$
    – Claude Leibovici
    Dec 9 '18 at 6:30










  • $begingroup$
    You should first try manipulating the given expression to make use of standard limits. Most limit problems on this website don't need advanced tools like L'Hospital's Rule or Taylor. Even when these tools are needed the prior manipulation greatly simplifies the use of these tools.
    $endgroup$
    – Paramanand Singh
    Dec 9 '18 at 14:01


















  • $begingroup$
    Do you have a specific example? You're generalising too much when there are a lot of limits you could simplify with L'Hopital's or Taylor series.
    $endgroup$
    – Toby Mak
    Dec 9 '18 at 5:29












  • $begingroup$
    @TobyMak, I meant the limits where L'Hospital Rule cannot be applied. I'll add an example in a few minutes. I just wanted to know if there are certain situations where we should use Taylor Expansions.
    $endgroup$
    – Harshil Bhatt
    Dec 9 '18 at 5:34












  • $begingroup$
    My point still holds. Generally speaking, if there are trigonometric functions such as $sin x, cos x, ln x$, or exponential functions such as $e^x$ and $ln x$, then the Taylor expansions will be easier.
    $endgroup$
    – Toby Mak
    Dec 9 '18 at 5:36












  • $begingroup$
    Have a look to this small nightmare : math.stackexchange.com/questions/925916/…
    $endgroup$
    – Claude Leibovici
    Dec 9 '18 at 6:30










  • $begingroup$
    You should first try manipulating the given expression to make use of standard limits. Most limit problems on this website don't need advanced tools like L'Hospital's Rule or Taylor. Even when these tools are needed the prior manipulation greatly simplifies the use of these tools.
    $endgroup$
    – Paramanand Singh
    Dec 9 '18 at 14:01
















$begingroup$
Do you have a specific example? You're generalising too much when there are a lot of limits you could simplify with L'Hopital's or Taylor series.
$endgroup$
– Toby Mak
Dec 9 '18 at 5:29






$begingroup$
Do you have a specific example? You're generalising too much when there are a lot of limits you could simplify with L'Hopital's or Taylor series.
$endgroup$
– Toby Mak
Dec 9 '18 at 5:29














$begingroup$
@TobyMak, I meant the limits where L'Hospital Rule cannot be applied. I'll add an example in a few minutes. I just wanted to know if there are certain situations where we should use Taylor Expansions.
$endgroup$
– Harshil Bhatt
Dec 9 '18 at 5:34






$begingroup$
@TobyMak, I meant the limits where L'Hospital Rule cannot be applied. I'll add an example in a few minutes. I just wanted to know if there are certain situations where we should use Taylor Expansions.
$endgroup$
– Harshil Bhatt
Dec 9 '18 at 5:34














$begingroup$
My point still holds. Generally speaking, if there are trigonometric functions such as $sin x, cos x, ln x$, or exponential functions such as $e^x$ and $ln x$, then the Taylor expansions will be easier.
$endgroup$
– Toby Mak
Dec 9 '18 at 5:36






$begingroup$
My point still holds. Generally speaking, if there are trigonometric functions such as $sin x, cos x, ln x$, or exponential functions such as $e^x$ and $ln x$, then the Taylor expansions will be easier.
$endgroup$
– Toby Mak
Dec 9 '18 at 5:36














$begingroup$
Have a look to this small nightmare : math.stackexchange.com/questions/925916/…
$endgroup$
– Claude Leibovici
Dec 9 '18 at 6:30




$begingroup$
Have a look to this small nightmare : math.stackexchange.com/questions/925916/…
$endgroup$
– Claude Leibovici
Dec 9 '18 at 6:30












$begingroup$
You should first try manipulating the given expression to make use of standard limits. Most limit problems on this website don't need advanced tools like L'Hospital's Rule or Taylor. Even when these tools are needed the prior manipulation greatly simplifies the use of these tools.
$endgroup$
– Paramanand Singh
Dec 9 '18 at 14:01




$begingroup$
You should first try manipulating the given expression to make use of standard limits. Most limit problems on this website don't need advanced tools like L'Hospital's Rule or Taylor. Even when these tools are needed the prior manipulation greatly simplifies the use of these tools.
$endgroup$
– Paramanand Singh
Dec 9 '18 at 14:01










3 Answers
3






active

oldest

votes


















2












$begingroup$

A general rule to follow is that we always should try at first to apply the simplest method for any limit. Therefore, when we face with an indeterminate form, I suggest at first to try with:



1. Algebraic manipulation, aimed to eliminate the source of that indetermination, e.g.



$$lim_{xto 1} frac{x^2-3x+2}{x^2-1}=lim_{xto 1} frac{(x-1)(x-2)}{(x-1)(x+1)}=lim_{xto 1} frac{x-2}{x+1}=-frac12$$



2. Standard limits, e.g.



$$xto 0, quad frac{sin x}xto 1, quad frac{e^x-1}{x}to 1, quad xlog x to 0,quad ldots$$



or related results derived from them, see for example: Are all limits solvable without L'Hôpital Rule or Series Expansion.



When we can't solve a limit in an indeterminate form by that basic tools, then we are allowed to use more advanced methods which involve derivatitive concepts and notably l'Hopital rule and Taylor's expansion.



In my opinion, with some special exception and when we are not explicitly requested/forced to use l'Hopital, we always should prefer Taylor's expansion not only because it is a more powerful and effective method but also because it allows us to really understand what is going on and which are the terms which really count in the expression whereas l'Hopital is a blind/black box method which doesn't add any contribution to our knowledge about limits.



Therefore if you know Taylor's expansion, in general, I suggest to proceed always by that.




Edit



The given example is one of the first category that is "solvable by elementary methods".



Notably we have that as $x to 0^+$ (otherwise the expression in not defined)



$$left(sin xright)^{frac{1}{x}}+left(frac{1}{x}right)^{sin x}=left(sin xright)^{frac{1}{x}}+frac{1}{x^{sin x}}to 0 +1=1$$



Indeed we have that




  • $left(sin xright)^{frac{1}{x}}to 0$ (it is in a not-indeterminate form $0^{infty}$)


  • $x^{sin x}=e^{sin x log x}to e^0=1$



since by standard limits



$$sin x log x= frac{sin x}xcdot xlog xto 1 cdot 0 =0$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But in an examination scenario, when time is of utmost importance, L'Hospital Rule would be the best way.
    $endgroup$
    – Harshil Bhatt
    Dec 11 '18 at 17:11










  • $begingroup$
    @HarshilBhatt Of course in some cases it can be useful to check the result for simple limit but, in generl, I suggest to do not learn limits by l'Hopital rule, it is really not helpful. In general, for limit not solveble by elementary methods, once we know Taylor's expansion, that's the preferable and more effective way to solve limits.
    $endgroup$
    – gimusi
    Dec 11 '18 at 17:18










  • $begingroup$
    @HarshilBhatt Sorry I didn't take a close look to that but the given example can be of course solved by standard limit. I add something on that.
    $endgroup$
    – gimusi
    Dec 11 '18 at 17:27










  • $begingroup$
    @HarshilBhatt: in a competitive exam scenario Taylor is far more effective provided you are comfortable with manipulation of series (like multiplying, dividing and composing them).
    $endgroup$
    – Paramanand Singh
    Dec 12 '18 at 1:05





















1












$begingroup$

I should say that most of the time (not to say all the times) Taylor expansions are the keys.



Let me consider your example
$$lim _{xto 0}left(left(sin(x)right)^{frac{1}{x}}+left(frac{1}{x}right)^{sin(x)}right)$$



Consider the first term
$$a=left(sin(x)right)^{frac{1}{x}} implies log(a)={frac{1}{x}}logleft(sin(x)right)$$ Now, by Taylor
$$sin(x)=x-frac{x^3}{6}+Oleft(x^5right)$$
$$logleft(sin(x)right)=log (x)-frac{x^2}{6}+Oleft(x^4right)$$
$$log(a)=frac{log (x)}{x}-frac{x}{6}+Oleft(x^3right)$$ So $log(a)to -infty$ which means thet $ato 0$.



Now, the second term (using equivalent is faster here)
$$b=left(frac{1}{x}right)^{sin(x)}implies log(b)=-sin(x) log(x)sim -x log(x)to 0^+$$ which means that $bto 1^+$ and so $a+b$.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    When you see, as $fto0$, the following functions, it is often a good way to use Taylor series rather than L'Hospital rule.




    • $log(1+f) = f+o(f)$

    • $sin f = f+o(f)$

    • $cos f = 1+ o(1)$

    • $e^f = 1+o(1)$

    • $tan f = f+o(f)$

    • $arctan f = f+o(f)$


    (More terms of the series are sometimes necessary).






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      What you are indicating are the first order expansion which can be proved solely by standard limits and do no requres derivative concept. Moreover when we deal with limits I always suggest to include the remainder term, that is $log(1+x)=x+o(x)$ etc., the asympthotic notation is really dangerous.
      $endgroup$
      – gimusi
      Dec 12 '18 at 7:33










    • $begingroup$
      Moreover your suggestion, "when we see...then" is not a good guide in general. The first try should always be done with simpler method (algebraic manipulation, standard limits).
      $endgroup$
      – gimusi
      Dec 12 '18 at 7:35










    • $begingroup$
      @gimusi I would say using Taylor series is quite a simple method, once you understand the meaning of it. I just wrote down the first order expansions to give an example; as I said, more terms of the series are sometimes necessary.
      $endgroup$
      – Lorenzo B.
      Dec 12 '18 at 14:37










    • $begingroup$
      Yes I agree with you, Taylor's series is of course the most powerfull and effective tool for limits but, for an educational purpouse, in my opinion at first we should always try to use elementary methods whenever possible. For the order of the expansions my main concern was for the remainder which I suggest to include always when dealing with limits. Bye
      $endgroup$
      – gimusi
      Dec 12 '18 at 14:47








    • 1




      $begingroup$
      That's was exactly my main point. Thanks, Bye
      $endgroup$
      – gimusi
      Dec 12 '18 at 15:02











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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    A general rule to follow is that we always should try at first to apply the simplest method for any limit. Therefore, when we face with an indeterminate form, I suggest at first to try with:



    1. Algebraic manipulation, aimed to eliminate the source of that indetermination, e.g.



    $$lim_{xto 1} frac{x^2-3x+2}{x^2-1}=lim_{xto 1} frac{(x-1)(x-2)}{(x-1)(x+1)}=lim_{xto 1} frac{x-2}{x+1}=-frac12$$



    2. Standard limits, e.g.



    $$xto 0, quad frac{sin x}xto 1, quad frac{e^x-1}{x}to 1, quad xlog x to 0,quad ldots$$



    or related results derived from them, see for example: Are all limits solvable without L'Hôpital Rule or Series Expansion.



    When we can't solve a limit in an indeterminate form by that basic tools, then we are allowed to use more advanced methods which involve derivatitive concepts and notably l'Hopital rule and Taylor's expansion.



    In my opinion, with some special exception and when we are not explicitly requested/forced to use l'Hopital, we always should prefer Taylor's expansion not only because it is a more powerful and effective method but also because it allows us to really understand what is going on and which are the terms which really count in the expression whereas l'Hopital is a blind/black box method which doesn't add any contribution to our knowledge about limits.



    Therefore if you know Taylor's expansion, in general, I suggest to proceed always by that.




    Edit



    The given example is one of the first category that is "solvable by elementary methods".



    Notably we have that as $x to 0^+$ (otherwise the expression in not defined)



    $$left(sin xright)^{frac{1}{x}}+left(frac{1}{x}right)^{sin x}=left(sin xright)^{frac{1}{x}}+frac{1}{x^{sin x}}to 0 +1=1$$



    Indeed we have that




    • $left(sin xright)^{frac{1}{x}}to 0$ (it is in a not-indeterminate form $0^{infty}$)


    • $x^{sin x}=e^{sin x log x}to e^0=1$



    since by standard limits



    $$sin x log x= frac{sin x}xcdot xlog xto 1 cdot 0 =0$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      But in an examination scenario, when time is of utmost importance, L'Hospital Rule would be the best way.
      $endgroup$
      – Harshil Bhatt
      Dec 11 '18 at 17:11










    • $begingroup$
      @HarshilBhatt Of course in some cases it can be useful to check the result for simple limit but, in generl, I suggest to do not learn limits by l'Hopital rule, it is really not helpful. In general, for limit not solveble by elementary methods, once we know Taylor's expansion, that's the preferable and more effective way to solve limits.
      $endgroup$
      – gimusi
      Dec 11 '18 at 17:18










    • $begingroup$
      @HarshilBhatt Sorry I didn't take a close look to that but the given example can be of course solved by standard limit. I add something on that.
      $endgroup$
      – gimusi
      Dec 11 '18 at 17:27










    • $begingroup$
      @HarshilBhatt: in a competitive exam scenario Taylor is far more effective provided you are comfortable with manipulation of series (like multiplying, dividing and composing them).
      $endgroup$
      – Paramanand Singh
      Dec 12 '18 at 1:05


















    2












    $begingroup$

    A general rule to follow is that we always should try at first to apply the simplest method for any limit. Therefore, when we face with an indeterminate form, I suggest at first to try with:



    1. Algebraic manipulation, aimed to eliminate the source of that indetermination, e.g.



    $$lim_{xto 1} frac{x^2-3x+2}{x^2-1}=lim_{xto 1} frac{(x-1)(x-2)}{(x-1)(x+1)}=lim_{xto 1} frac{x-2}{x+1}=-frac12$$



    2. Standard limits, e.g.



    $$xto 0, quad frac{sin x}xto 1, quad frac{e^x-1}{x}to 1, quad xlog x to 0,quad ldots$$



    or related results derived from them, see for example: Are all limits solvable without L'Hôpital Rule or Series Expansion.



    When we can't solve a limit in an indeterminate form by that basic tools, then we are allowed to use more advanced methods which involve derivatitive concepts and notably l'Hopital rule and Taylor's expansion.



    In my opinion, with some special exception and when we are not explicitly requested/forced to use l'Hopital, we always should prefer Taylor's expansion not only because it is a more powerful and effective method but also because it allows us to really understand what is going on and which are the terms which really count in the expression whereas l'Hopital is a blind/black box method which doesn't add any contribution to our knowledge about limits.



    Therefore if you know Taylor's expansion, in general, I suggest to proceed always by that.




    Edit



    The given example is one of the first category that is "solvable by elementary methods".



    Notably we have that as $x to 0^+$ (otherwise the expression in not defined)



    $$left(sin xright)^{frac{1}{x}}+left(frac{1}{x}right)^{sin x}=left(sin xright)^{frac{1}{x}}+frac{1}{x^{sin x}}to 0 +1=1$$



    Indeed we have that




    • $left(sin xright)^{frac{1}{x}}to 0$ (it is in a not-indeterminate form $0^{infty}$)


    • $x^{sin x}=e^{sin x log x}to e^0=1$



    since by standard limits



    $$sin x log x= frac{sin x}xcdot xlog xto 1 cdot 0 =0$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      But in an examination scenario, when time is of utmost importance, L'Hospital Rule would be the best way.
      $endgroup$
      – Harshil Bhatt
      Dec 11 '18 at 17:11










    • $begingroup$
      @HarshilBhatt Of course in some cases it can be useful to check the result for simple limit but, in generl, I suggest to do not learn limits by l'Hopital rule, it is really not helpful. In general, for limit not solveble by elementary methods, once we know Taylor's expansion, that's the preferable and more effective way to solve limits.
      $endgroup$
      – gimusi
      Dec 11 '18 at 17:18










    • $begingroup$
      @HarshilBhatt Sorry I didn't take a close look to that but the given example can be of course solved by standard limit. I add something on that.
      $endgroup$
      – gimusi
      Dec 11 '18 at 17:27










    • $begingroup$
      @HarshilBhatt: in a competitive exam scenario Taylor is far more effective provided you are comfortable with manipulation of series (like multiplying, dividing and composing them).
      $endgroup$
      – Paramanand Singh
      Dec 12 '18 at 1:05
















    2












    2








    2





    $begingroup$

    A general rule to follow is that we always should try at first to apply the simplest method for any limit. Therefore, when we face with an indeterminate form, I suggest at first to try with:



    1. Algebraic manipulation, aimed to eliminate the source of that indetermination, e.g.



    $$lim_{xto 1} frac{x^2-3x+2}{x^2-1}=lim_{xto 1} frac{(x-1)(x-2)}{(x-1)(x+1)}=lim_{xto 1} frac{x-2}{x+1}=-frac12$$



    2. Standard limits, e.g.



    $$xto 0, quad frac{sin x}xto 1, quad frac{e^x-1}{x}to 1, quad xlog x to 0,quad ldots$$



    or related results derived from them, see for example: Are all limits solvable without L'Hôpital Rule or Series Expansion.



    When we can't solve a limit in an indeterminate form by that basic tools, then we are allowed to use more advanced methods which involve derivatitive concepts and notably l'Hopital rule and Taylor's expansion.



    In my opinion, with some special exception and when we are not explicitly requested/forced to use l'Hopital, we always should prefer Taylor's expansion not only because it is a more powerful and effective method but also because it allows us to really understand what is going on and which are the terms which really count in the expression whereas l'Hopital is a blind/black box method which doesn't add any contribution to our knowledge about limits.



    Therefore if you know Taylor's expansion, in general, I suggest to proceed always by that.




    Edit



    The given example is one of the first category that is "solvable by elementary methods".



    Notably we have that as $x to 0^+$ (otherwise the expression in not defined)



    $$left(sin xright)^{frac{1}{x}}+left(frac{1}{x}right)^{sin x}=left(sin xright)^{frac{1}{x}}+frac{1}{x^{sin x}}to 0 +1=1$$



    Indeed we have that




    • $left(sin xright)^{frac{1}{x}}to 0$ (it is in a not-indeterminate form $0^{infty}$)


    • $x^{sin x}=e^{sin x log x}to e^0=1$



    since by standard limits



    $$sin x log x= frac{sin x}xcdot xlog xto 1 cdot 0 =0$$






    share|cite|improve this answer











    $endgroup$



    A general rule to follow is that we always should try at first to apply the simplest method for any limit. Therefore, when we face with an indeterminate form, I suggest at first to try with:



    1. Algebraic manipulation, aimed to eliminate the source of that indetermination, e.g.



    $$lim_{xto 1} frac{x^2-3x+2}{x^2-1}=lim_{xto 1} frac{(x-1)(x-2)}{(x-1)(x+1)}=lim_{xto 1} frac{x-2}{x+1}=-frac12$$



    2. Standard limits, e.g.



    $$xto 0, quad frac{sin x}xto 1, quad frac{e^x-1}{x}to 1, quad xlog x to 0,quad ldots$$



    or related results derived from them, see for example: Are all limits solvable without L'Hôpital Rule or Series Expansion.



    When we can't solve a limit in an indeterminate form by that basic tools, then we are allowed to use more advanced methods which involve derivatitive concepts and notably l'Hopital rule and Taylor's expansion.



    In my opinion, with some special exception and when we are not explicitly requested/forced to use l'Hopital, we always should prefer Taylor's expansion not only because it is a more powerful and effective method but also because it allows us to really understand what is going on and which are the terms which really count in the expression whereas l'Hopital is a blind/black box method which doesn't add any contribution to our knowledge about limits.



    Therefore if you know Taylor's expansion, in general, I suggest to proceed always by that.




    Edit



    The given example is one of the first category that is "solvable by elementary methods".



    Notably we have that as $x to 0^+$ (otherwise the expression in not defined)



    $$left(sin xright)^{frac{1}{x}}+left(frac{1}{x}right)^{sin x}=left(sin xright)^{frac{1}{x}}+frac{1}{x^{sin x}}to 0 +1=1$$



    Indeed we have that




    • $left(sin xright)^{frac{1}{x}}to 0$ (it is in a not-indeterminate form $0^{infty}$)


    • $x^{sin x}=e^{sin x log x}to e^0=1$



    since by standard limits



    $$sin x log x= frac{sin x}xcdot xlog xto 1 cdot 0 =0$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 11 '18 at 17:38

























    answered Dec 11 '18 at 12:57









    gimusigimusi

    92.8k84494




    92.8k84494












    • $begingroup$
      But in an examination scenario, when time is of utmost importance, L'Hospital Rule would be the best way.
      $endgroup$
      – Harshil Bhatt
      Dec 11 '18 at 17:11










    • $begingroup$
      @HarshilBhatt Of course in some cases it can be useful to check the result for simple limit but, in generl, I suggest to do not learn limits by l'Hopital rule, it is really not helpful. In general, for limit not solveble by elementary methods, once we know Taylor's expansion, that's the preferable and more effective way to solve limits.
      $endgroup$
      – gimusi
      Dec 11 '18 at 17:18










    • $begingroup$
      @HarshilBhatt Sorry I didn't take a close look to that but the given example can be of course solved by standard limit. I add something on that.
      $endgroup$
      – gimusi
      Dec 11 '18 at 17:27










    • $begingroup$
      @HarshilBhatt: in a competitive exam scenario Taylor is far more effective provided you are comfortable with manipulation of series (like multiplying, dividing and composing them).
      $endgroup$
      – Paramanand Singh
      Dec 12 '18 at 1:05




















    • $begingroup$
      But in an examination scenario, when time is of utmost importance, L'Hospital Rule would be the best way.
      $endgroup$
      – Harshil Bhatt
      Dec 11 '18 at 17:11










    • $begingroup$
      @HarshilBhatt Of course in some cases it can be useful to check the result for simple limit but, in generl, I suggest to do not learn limits by l'Hopital rule, it is really not helpful. In general, for limit not solveble by elementary methods, once we know Taylor's expansion, that's the preferable and more effective way to solve limits.
      $endgroup$
      – gimusi
      Dec 11 '18 at 17:18










    • $begingroup$
      @HarshilBhatt Sorry I didn't take a close look to that but the given example can be of course solved by standard limit. I add something on that.
      $endgroup$
      – gimusi
      Dec 11 '18 at 17:27










    • $begingroup$
      @HarshilBhatt: in a competitive exam scenario Taylor is far more effective provided you are comfortable with manipulation of series (like multiplying, dividing and composing them).
      $endgroup$
      – Paramanand Singh
      Dec 12 '18 at 1:05


















    $begingroup$
    But in an examination scenario, when time is of utmost importance, L'Hospital Rule would be the best way.
    $endgroup$
    – Harshil Bhatt
    Dec 11 '18 at 17:11




    $begingroup$
    But in an examination scenario, when time is of utmost importance, L'Hospital Rule would be the best way.
    $endgroup$
    – Harshil Bhatt
    Dec 11 '18 at 17:11












    $begingroup$
    @HarshilBhatt Of course in some cases it can be useful to check the result for simple limit but, in generl, I suggest to do not learn limits by l'Hopital rule, it is really not helpful. In general, for limit not solveble by elementary methods, once we know Taylor's expansion, that's the preferable and more effective way to solve limits.
    $endgroup$
    – gimusi
    Dec 11 '18 at 17:18




    $begingroup$
    @HarshilBhatt Of course in some cases it can be useful to check the result for simple limit but, in generl, I suggest to do not learn limits by l'Hopital rule, it is really not helpful. In general, for limit not solveble by elementary methods, once we know Taylor's expansion, that's the preferable and more effective way to solve limits.
    $endgroup$
    – gimusi
    Dec 11 '18 at 17:18












    $begingroup$
    @HarshilBhatt Sorry I didn't take a close look to that but the given example can be of course solved by standard limit. I add something on that.
    $endgroup$
    – gimusi
    Dec 11 '18 at 17:27




    $begingroup$
    @HarshilBhatt Sorry I didn't take a close look to that but the given example can be of course solved by standard limit. I add something on that.
    $endgroup$
    – gimusi
    Dec 11 '18 at 17:27












    $begingroup$
    @HarshilBhatt: in a competitive exam scenario Taylor is far more effective provided you are comfortable with manipulation of series (like multiplying, dividing and composing them).
    $endgroup$
    – Paramanand Singh
    Dec 12 '18 at 1:05






    $begingroup$
    @HarshilBhatt: in a competitive exam scenario Taylor is far more effective provided you are comfortable with manipulation of series (like multiplying, dividing and composing them).
    $endgroup$
    – Paramanand Singh
    Dec 12 '18 at 1:05













    1












    $begingroup$

    I should say that most of the time (not to say all the times) Taylor expansions are the keys.



    Let me consider your example
    $$lim _{xto 0}left(left(sin(x)right)^{frac{1}{x}}+left(frac{1}{x}right)^{sin(x)}right)$$



    Consider the first term
    $$a=left(sin(x)right)^{frac{1}{x}} implies log(a)={frac{1}{x}}logleft(sin(x)right)$$ Now, by Taylor
    $$sin(x)=x-frac{x^3}{6}+Oleft(x^5right)$$
    $$logleft(sin(x)right)=log (x)-frac{x^2}{6}+Oleft(x^4right)$$
    $$log(a)=frac{log (x)}{x}-frac{x}{6}+Oleft(x^3right)$$ So $log(a)to -infty$ which means thet $ato 0$.



    Now, the second term (using equivalent is faster here)
    $$b=left(frac{1}{x}right)^{sin(x)}implies log(b)=-sin(x) log(x)sim -x log(x)to 0^+$$ which means that $bto 1^+$ and so $a+b$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      I should say that most of the time (not to say all the times) Taylor expansions are the keys.



      Let me consider your example
      $$lim _{xto 0}left(left(sin(x)right)^{frac{1}{x}}+left(frac{1}{x}right)^{sin(x)}right)$$



      Consider the first term
      $$a=left(sin(x)right)^{frac{1}{x}} implies log(a)={frac{1}{x}}logleft(sin(x)right)$$ Now, by Taylor
      $$sin(x)=x-frac{x^3}{6}+Oleft(x^5right)$$
      $$logleft(sin(x)right)=log (x)-frac{x^2}{6}+Oleft(x^4right)$$
      $$log(a)=frac{log (x)}{x}-frac{x}{6}+Oleft(x^3right)$$ So $log(a)to -infty$ which means thet $ato 0$.



      Now, the second term (using equivalent is faster here)
      $$b=left(frac{1}{x}right)^{sin(x)}implies log(b)=-sin(x) log(x)sim -x log(x)to 0^+$$ which means that $bto 1^+$ and so $a+b$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        I should say that most of the time (not to say all the times) Taylor expansions are the keys.



        Let me consider your example
        $$lim _{xto 0}left(left(sin(x)right)^{frac{1}{x}}+left(frac{1}{x}right)^{sin(x)}right)$$



        Consider the first term
        $$a=left(sin(x)right)^{frac{1}{x}} implies log(a)={frac{1}{x}}logleft(sin(x)right)$$ Now, by Taylor
        $$sin(x)=x-frac{x^3}{6}+Oleft(x^5right)$$
        $$logleft(sin(x)right)=log (x)-frac{x^2}{6}+Oleft(x^4right)$$
        $$log(a)=frac{log (x)}{x}-frac{x}{6}+Oleft(x^3right)$$ So $log(a)to -infty$ which means thet $ato 0$.



        Now, the second term (using equivalent is faster here)
        $$b=left(frac{1}{x}right)^{sin(x)}implies log(b)=-sin(x) log(x)sim -x log(x)to 0^+$$ which means that $bto 1^+$ and so $a+b$.






        share|cite|improve this answer











        $endgroup$



        I should say that most of the time (not to say all the times) Taylor expansions are the keys.



        Let me consider your example
        $$lim _{xto 0}left(left(sin(x)right)^{frac{1}{x}}+left(frac{1}{x}right)^{sin(x)}right)$$



        Consider the first term
        $$a=left(sin(x)right)^{frac{1}{x}} implies log(a)={frac{1}{x}}logleft(sin(x)right)$$ Now, by Taylor
        $$sin(x)=x-frac{x^3}{6}+Oleft(x^5right)$$
        $$logleft(sin(x)right)=log (x)-frac{x^2}{6}+Oleft(x^4right)$$
        $$log(a)=frac{log (x)}{x}-frac{x}{6}+Oleft(x^3right)$$ So $log(a)to -infty$ which means thet $ato 0$.



        Now, the second term (using equivalent is faster here)
        $$b=left(frac{1}{x}right)^{sin(x)}implies log(b)=-sin(x) log(x)sim -x log(x)to 0^+$$ which means that $bto 1^+$ and so $a+b$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 9 '18 at 6:59

























        answered Dec 9 '18 at 6:18









        Claude LeiboviciClaude Leibovici

        120k1157132




        120k1157132























            1












            $begingroup$

            When you see, as $fto0$, the following functions, it is often a good way to use Taylor series rather than L'Hospital rule.




            • $log(1+f) = f+o(f)$

            • $sin f = f+o(f)$

            • $cos f = 1+ o(1)$

            • $e^f = 1+o(1)$

            • $tan f = f+o(f)$

            • $arctan f = f+o(f)$


            (More terms of the series are sometimes necessary).






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              What you are indicating are the first order expansion which can be proved solely by standard limits and do no requres derivative concept. Moreover when we deal with limits I always suggest to include the remainder term, that is $log(1+x)=x+o(x)$ etc., the asympthotic notation is really dangerous.
              $endgroup$
              – gimusi
              Dec 12 '18 at 7:33










            • $begingroup$
              Moreover your suggestion, "when we see...then" is not a good guide in general. The first try should always be done with simpler method (algebraic manipulation, standard limits).
              $endgroup$
              – gimusi
              Dec 12 '18 at 7:35










            • $begingroup$
              @gimusi I would say using Taylor series is quite a simple method, once you understand the meaning of it. I just wrote down the first order expansions to give an example; as I said, more terms of the series are sometimes necessary.
              $endgroup$
              – Lorenzo B.
              Dec 12 '18 at 14:37










            • $begingroup$
              Yes I agree with you, Taylor's series is of course the most powerfull and effective tool for limits but, for an educational purpouse, in my opinion at first we should always try to use elementary methods whenever possible. For the order of the expansions my main concern was for the remainder which I suggest to include always when dealing with limits. Bye
              $endgroup$
              – gimusi
              Dec 12 '18 at 14:47








            • 1




              $begingroup$
              That's was exactly my main point. Thanks, Bye
              $endgroup$
              – gimusi
              Dec 12 '18 at 15:02
















            1












            $begingroup$

            When you see, as $fto0$, the following functions, it is often a good way to use Taylor series rather than L'Hospital rule.




            • $log(1+f) = f+o(f)$

            • $sin f = f+o(f)$

            • $cos f = 1+ o(1)$

            • $e^f = 1+o(1)$

            • $tan f = f+o(f)$

            • $arctan f = f+o(f)$


            (More terms of the series are sometimes necessary).






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              What you are indicating are the first order expansion which can be proved solely by standard limits and do no requres derivative concept. Moreover when we deal with limits I always suggest to include the remainder term, that is $log(1+x)=x+o(x)$ etc., the asympthotic notation is really dangerous.
              $endgroup$
              – gimusi
              Dec 12 '18 at 7:33










            • $begingroup$
              Moreover your suggestion, "when we see...then" is not a good guide in general. The first try should always be done with simpler method (algebraic manipulation, standard limits).
              $endgroup$
              – gimusi
              Dec 12 '18 at 7:35










            • $begingroup$
              @gimusi I would say using Taylor series is quite a simple method, once you understand the meaning of it. I just wrote down the first order expansions to give an example; as I said, more terms of the series are sometimes necessary.
              $endgroup$
              – Lorenzo B.
              Dec 12 '18 at 14:37










            • $begingroup$
              Yes I agree with you, Taylor's series is of course the most powerfull and effective tool for limits but, for an educational purpouse, in my opinion at first we should always try to use elementary methods whenever possible. For the order of the expansions my main concern was for the remainder which I suggest to include always when dealing with limits. Bye
              $endgroup$
              – gimusi
              Dec 12 '18 at 14:47








            • 1




              $begingroup$
              That's was exactly my main point. Thanks, Bye
              $endgroup$
              – gimusi
              Dec 12 '18 at 15:02














            1












            1








            1





            $begingroup$

            When you see, as $fto0$, the following functions, it is often a good way to use Taylor series rather than L'Hospital rule.




            • $log(1+f) = f+o(f)$

            • $sin f = f+o(f)$

            • $cos f = 1+ o(1)$

            • $e^f = 1+o(1)$

            • $tan f = f+o(f)$

            • $arctan f = f+o(f)$


            (More terms of the series are sometimes necessary).






            share|cite|improve this answer











            $endgroup$



            When you see, as $fto0$, the following functions, it is often a good way to use Taylor series rather than L'Hospital rule.




            • $log(1+f) = f+o(f)$

            • $sin f = f+o(f)$

            • $cos f = 1+ o(1)$

            • $e^f = 1+o(1)$

            • $tan f = f+o(f)$

            • $arctan f = f+o(f)$


            (More terms of the series are sometimes necessary).







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 12 '18 at 14:32

























            answered Dec 11 '18 at 19:50









            Lorenzo B.Lorenzo B.

            1,8402520




            1,8402520












            • $begingroup$
              What you are indicating are the first order expansion which can be proved solely by standard limits and do no requres derivative concept. Moreover when we deal with limits I always suggest to include the remainder term, that is $log(1+x)=x+o(x)$ etc., the asympthotic notation is really dangerous.
              $endgroup$
              – gimusi
              Dec 12 '18 at 7:33










            • $begingroup$
              Moreover your suggestion, "when we see...then" is not a good guide in general. The first try should always be done with simpler method (algebraic manipulation, standard limits).
              $endgroup$
              – gimusi
              Dec 12 '18 at 7:35










            • $begingroup$
              @gimusi I would say using Taylor series is quite a simple method, once you understand the meaning of it. I just wrote down the first order expansions to give an example; as I said, more terms of the series are sometimes necessary.
              $endgroup$
              – Lorenzo B.
              Dec 12 '18 at 14:37










            • $begingroup$
              Yes I agree with you, Taylor's series is of course the most powerfull and effective tool for limits but, for an educational purpouse, in my opinion at first we should always try to use elementary methods whenever possible. For the order of the expansions my main concern was for the remainder which I suggest to include always when dealing with limits. Bye
              $endgroup$
              – gimusi
              Dec 12 '18 at 14:47








            • 1




              $begingroup$
              That's was exactly my main point. Thanks, Bye
              $endgroup$
              – gimusi
              Dec 12 '18 at 15:02


















            • $begingroup$
              What you are indicating are the first order expansion which can be proved solely by standard limits and do no requres derivative concept. Moreover when we deal with limits I always suggest to include the remainder term, that is $log(1+x)=x+o(x)$ etc., the asympthotic notation is really dangerous.
              $endgroup$
              – gimusi
              Dec 12 '18 at 7:33










            • $begingroup$
              Moreover your suggestion, "when we see...then" is not a good guide in general. The first try should always be done with simpler method (algebraic manipulation, standard limits).
              $endgroup$
              – gimusi
              Dec 12 '18 at 7:35










            • $begingroup$
              @gimusi I would say using Taylor series is quite a simple method, once you understand the meaning of it. I just wrote down the first order expansions to give an example; as I said, more terms of the series are sometimes necessary.
              $endgroup$
              – Lorenzo B.
              Dec 12 '18 at 14:37










            • $begingroup$
              Yes I agree with you, Taylor's series is of course the most powerfull and effective tool for limits but, for an educational purpouse, in my opinion at first we should always try to use elementary methods whenever possible. For the order of the expansions my main concern was for the remainder which I suggest to include always when dealing with limits. Bye
              $endgroup$
              – gimusi
              Dec 12 '18 at 14:47








            • 1




              $begingroup$
              That's was exactly my main point. Thanks, Bye
              $endgroup$
              – gimusi
              Dec 12 '18 at 15:02
















            $begingroup$
            What you are indicating are the first order expansion which can be proved solely by standard limits and do no requres derivative concept. Moreover when we deal with limits I always suggest to include the remainder term, that is $log(1+x)=x+o(x)$ etc., the asympthotic notation is really dangerous.
            $endgroup$
            – gimusi
            Dec 12 '18 at 7:33




            $begingroup$
            What you are indicating are the first order expansion which can be proved solely by standard limits and do no requres derivative concept. Moreover when we deal with limits I always suggest to include the remainder term, that is $log(1+x)=x+o(x)$ etc., the asympthotic notation is really dangerous.
            $endgroup$
            – gimusi
            Dec 12 '18 at 7:33












            $begingroup$
            Moreover your suggestion, "when we see...then" is not a good guide in general. The first try should always be done with simpler method (algebraic manipulation, standard limits).
            $endgroup$
            – gimusi
            Dec 12 '18 at 7:35




            $begingroup$
            Moreover your suggestion, "when we see...then" is not a good guide in general. The first try should always be done with simpler method (algebraic manipulation, standard limits).
            $endgroup$
            – gimusi
            Dec 12 '18 at 7:35












            $begingroup$
            @gimusi I would say using Taylor series is quite a simple method, once you understand the meaning of it. I just wrote down the first order expansions to give an example; as I said, more terms of the series are sometimes necessary.
            $endgroup$
            – Lorenzo B.
            Dec 12 '18 at 14:37




            $begingroup$
            @gimusi I would say using Taylor series is quite a simple method, once you understand the meaning of it. I just wrote down the first order expansions to give an example; as I said, more terms of the series are sometimes necessary.
            $endgroup$
            – Lorenzo B.
            Dec 12 '18 at 14:37












            $begingroup$
            Yes I agree with you, Taylor's series is of course the most powerfull and effective tool for limits but, for an educational purpouse, in my opinion at first we should always try to use elementary methods whenever possible. For the order of the expansions my main concern was for the remainder which I suggest to include always when dealing with limits. Bye
            $endgroup$
            – gimusi
            Dec 12 '18 at 14:47






            $begingroup$
            Yes I agree with you, Taylor's series is of course the most powerfull and effective tool for limits but, for an educational purpouse, in my opinion at first we should always try to use elementary methods whenever possible. For the order of the expansions my main concern was for the remainder which I suggest to include always when dealing with limits. Bye
            $endgroup$
            – gimusi
            Dec 12 '18 at 14:47






            1




            1




            $begingroup$
            That's was exactly my main point. Thanks, Bye
            $endgroup$
            – gimusi
            Dec 12 '18 at 15:02




            $begingroup$
            That's was exactly my main point. Thanks, Bye
            $endgroup$
            – gimusi
            Dec 12 '18 at 15:02


















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            Mont Emei

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            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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