Limit of rational function solving












1












$begingroup$


I have to solve limit of rational function, but it turns out I do mistake somewhere. Where I do wrong? Does my calculations correct?



$$lim_{xto infty}frac{x^3-2x-1}{x^5-2x-1}$$
Step 1:
$$lim_{xtoinfty}frac{x^5left(frac{1}{x^2}-frac{2}{x^4}-frac{1}{x^5}right)}{x^5left(1-frac{2}{x^4}-frac{1}{x^5}right)}$$
Step 2:
$$lim_{xtoinfty}frac{x^5left(frac{1}{x^2}-frac{2}{x^4}-frac{1}{x^5}right)}{x^5left(1-frac{2}{x^4}-frac{1}{x^5}right)} = frac{0-0-0}{1-0-0}=frac{0}{1}=0$$






algebra-precalculus limits rational-functions







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  • $begingroup$
    So $1=1*infty$? I never knew that!
    $endgroup$
    – Lord Shark the Unknown
    May 25 '17 at 10:49










  • $begingroup$
    I cannot follow your every single step!!
    $endgroup$
    – velut luna
    May 25 '17 at 10:52










  • $begingroup$
    Please check edited answer. Is that correct?
    $endgroup$
    – TeodorKolev
    May 25 '17 at 10:56






  • 1




    $begingroup$
    The editing answer is exactly correct
    $endgroup$
    – Noor Aslam
    May 25 '17 at 11:02






  • 2




    $begingroup$
    The answer as it stands now certainly isn't correct, the numerator in step 1 has two incorrect terms, namely $-2/x^2$ and $-1/x^3$
    $endgroup$
    – B. Mehta
    May 25 '17 at 11:23
















1












$begingroup$


I have to solve limit of rational function, but it turns out I do mistake somewhere. Where I do wrong? Does my calculations correct?



$$lim_{xto infty}frac{x^3-2x-1}{x^5-2x-1}$$
Step 1:
$$lim_{xtoinfty}frac{x^5left(frac{1}{x^2}-frac{2}{x^4}-frac{1}{x^5}right)}{x^5left(1-frac{2}{x^4}-frac{1}{x^5}right)}$$
Step 2:
$$lim_{xtoinfty}frac{x^5left(frac{1}{x^2}-frac{2}{x^4}-frac{1}{x^5}right)}{x^5left(1-frac{2}{x^4}-frac{1}{x^5}right)} = frac{0-0-0}{1-0-0}=frac{0}{1}=0$$






algebra-precalculus limits rational-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    So $1=1*infty$? I never knew that!
    $endgroup$
    – Lord Shark the Unknown
    May 25 '17 at 10:49










  • $begingroup$
    I cannot follow your every single step!!
    $endgroup$
    – velut luna
    May 25 '17 at 10:52










  • $begingroup$
    Please check edited answer. Is that correct?
    $endgroup$
    – TeodorKolev
    May 25 '17 at 10:56






  • 1




    $begingroup$
    The editing answer is exactly correct
    $endgroup$
    – Noor Aslam
    May 25 '17 at 11:02






  • 2




    $begingroup$
    The answer as it stands now certainly isn't correct, the numerator in step 1 has two incorrect terms, namely $-2/x^2$ and $-1/x^3$
    $endgroup$
    – B. Mehta
    May 25 '17 at 11:23














1












1








1





$begingroup$


I have to solve limit of rational function, but it turns out I do mistake somewhere. Where I do wrong? Does my calculations correct?



$$lim_{xto infty}frac{x^3-2x-1}{x^5-2x-1}$$
Step 1:
$$lim_{xtoinfty}frac{x^5left(frac{1}{x^2}-frac{2}{x^4}-frac{1}{x^5}right)}{x^5left(1-frac{2}{x^4}-frac{1}{x^5}right)}$$
Step 2:
$$lim_{xtoinfty}frac{x^5left(frac{1}{x^2}-frac{2}{x^4}-frac{1}{x^5}right)}{x^5left(1-frac{2}{x^4}-frac{1}{x^5}right)} = frac{0-0-0}{1-0-0}=frac{0}{1}=0$$






algebra-precalculus limits rational-functions







share|cite|improve this question











$endgroup$




I have to solve limit of rational function, but it turns out I do mistake somewhere. Where I do wrong? Does my calculations correct?



$$lim_{xto infty}frac{x^3-2x-1}{x^5-2x-1}$$
Step 1:
$$lim_{xtoinfty}frac{x^5left(frac{1}{x^2}-frac{2}{x^4}-frac{1}{x^5}right)}{x^5left(1-frac{2}{x^4}-frac{1}{x^5}right)}$$
Step 2:
$$lim_{xtoinfty}frac{x^5left(frac{1}{x^2}-frac{2}{x^4}-frac{1}{x^5}right)}{x^5left(1-frac{2}{x^4}-frac{1}{x^5}right)} = frac{0-0-0}{1-0-0}=frac{0}{1}=0$$






algebra-precalculus limits rational-functions




algebra-precalculus limits rational-functions






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited May 25 '17 at 11:30







TeodorKolev

















asked May 25 '17 at 10:47









TeodorKolevTeodorKolev

207212




207212












  • $begingroup$
    So $1=1*infty$? I never knew that!
    $endgroup$
    – Lord Shark the Unknown
    May 25 '17 at 10:49










  • $begingroup$
    I cannot follow your every single step!!
    $endgroup$
    – velut luna
    May 25 '17 at 10:52










  • $begingroup$
    Please check edited answer. Is that correct?
    $endgroup$
    – TeodorKolev
    May 25 '17 at 10:56






  • 1




    $begingroup$
    The editing answer is exactly correct
    $endgroup$
    – Noor Aslam
    May 25 '17 at 11:02






  • 2




    $begingroup$
    The answer as it stands now certainly isn't correct, the numerator in step 1 has two incorrect terms, namely $-2/x^2$ and $-1/x^3$
    $endgroup$
    – B. Mehta
    May 25 '17 at 11:23


















  • $begingroup$
    So $1=1*infty$? I never knew that!
    $endgroup$
    – Lord Shark the Unknown
    May 25 '17 at 10:49










  • $begingroup$
    I cannot follow your every single step!!
    $endgroup$
    – velut luna
    May 25 '17 at 10:52










  • $begingroup$
    Please check edited answer. Is that correct?
    $endgroup$
    – TeodorKolev
    May 25 '17 at 10:56






  • 1




    $begingroup$
    The editing answer is exactly correct
    $endgroup$
    – Noor Aslam
    May 25 '17 at 11:02






  • 2




    $begingroup$
    The answer as it stands now certainly isn't correct, the numerator in step 1 has two incorrect terms, namely $-2/x^2$ and $-1/x^3$
    $endgroup$
    – B. Mehta
    May 25 '17 at 11:23
















$begingroup$
So $1=1*infty$? I never knew that!
$endgroup$
– Lord Shark the Unknown
May 25 '17 at 10:49




$begingroup$
So $1=1*infty$? I never knew that!
$endgroup$
– Lord Shark the Unknown
May 25 '17 at 10:49












$begingroup$
I cannot follow your every single step!!
$endgroup$
– velut luna
May 25 '17 at 10:52




$begingroup$
I cannot follow your every single step!!
$endgroup$
– velut luna
May 25 '17 at 10:52












$begingroup$
Please check edited answer. Is that correct?
$endgroup$
– TeodorKolev
May 25 '17 at 10:56




$begingroup$
Please check edited answer. Is that correct?
$endgroup$
– TeodorKolev
May 25 '17 at 10:56




1




1




$begingroup$
The editing answer is exactly correct
$endgroup$
– Noor Aslam
May 25 '17 at 11:02




$begingroup$
The editing answer is exactly correct
$endgroup$
– Noor Aslam
May 25 '17 at 11:02




2




2




$begingroup$
The answer as it stands now certainly isn't correct, the numerator in step 1 has two incorrect terms, namely $-2/x^2$ and $-1/x^3$
$endgroup$
– B. Mehta
May 25 '17 at 11:23




$begingroup$
The answer as it stands now certainly isn't correct, the numerator in step 1 has two incorrect terms, namely $-2/x^2$ and $-1/x^3$
$endgroup$
– B. Mehta
May 25 '17 at 11:23










1 Answer
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oldest

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$begingroup$

The post edit answer is correct. However, it can be obtained with little (or practically no) calculations at all by using the properties of end behaviour of polynomials.
The end behaviour of a polynomial function can be defined, informally, as the behaviour of the function's graph as it approaches $pm infty$.



For polynomial functions, the degree and the leading coefficient determine the end behaviour of the function. That means, as $x$ tends to infinity it is enough to analyse the leading term and coefficient to have an idea of what the graph will look like.



Limits are, on one hand, asking about the end behaviour. In your case, you have two polynomial functions that make up a rational function. Therefore we can combine the property of end behaviour of polynomials and the quotient rule of limits, $${limlimits_{x to a} frac{{fleft( x right)}}{{gleft( x right)}} }={ frac{{limlimits_{x to a} fleft( x right)}}{{limlimits_{x to a} gleft( x right)}},;;;}kern-0.3pt
{text{if};;limlimits_{x to a} gleft( x right) ne 0}$$ to quickly get the answer by considering $$lim_{xto infty} frac{x^3}{x^5} = lim_{x to infty} frac{1}{x^2} = frac{1}{infty} = 0$$ Of course, all this can be done in the head in just a matter of seconds.



To Sum It Up



Suppose you are given two rational functions $$f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+cdots+a_{1}x+a_{0} $$and $$g(x)=b_{m}x^{m}+b_{m-1}x^{m-1}+cdots+b_{1}x+b_{0}.$$

Then the value of $$lim_{xrightarrowinfty}frac{f(x)}{g(x)}$$ can be determined as follows:



$bullet$ If $n>m$ and the sign of $a_n$ is the same as that of $b_m$ (i.e. $frac{a_{n}}{b_{m}}$ is positive) then the limit is $infty$



$bullet$ If $n>m$ and the sign of $a_n$ is different from that of $b_m$, (i.e. $frac{a_{n}}{b_{m}}$ is negative) then the limit is $-infty$



$bullet$ If $n<m$ then the limit is $0$.



$bullet$ If $n=m$ then the limit is the value of $frac{a_{n}}{b_{m}}$.



Note that this method is only for limits to infinity






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    1












    $begingroup$

    The post edit answer is correct. However, it can be obtained with little (or practically no) calculations at all by using the properties of end behaviour of polynomials.
    The end behaviour of a polynomial function can be defined, informally, as the behaviour of the function's graph as it approaches $pm infty$.



    For polynomial functions, the degree and the leading coefficient determine the end behaviour of the function. That means, as $x$ tends to infinity it is enough to analyse the leading term and coefficient to have an idea of what the graph will look like.



    Limits are, on one hand, asking about the end behaviour. In your case, you have two polynomial functions that make up a rational function. Therefore we can combine the property of end behaviour of polynomials and the quotient rule of limits, $${limlimits_{x to a} frac{{fleft( x right)}}{{gleft( x right)}} }={ frac{{limlimits_{x to a} fleft( x right)}}{{limlimits_{x to a} gleft( x right)}},;;;}kern-0.3pt
    {text{if};;limlimits_{x to a} gleft( x right) ne 0}$$     to quickly get the answer by considering $$lim_{xto infty} frac{x^3}{x^5} = lim_{x to infty} frac{1}{x^2} = frac{1}{infty} = 0$$ Of course, all this can be done in the head in just a matter of seconds.



    To Sum It Up



    Suppose you are given two rational functions $$f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+cdots+a_{1}x+a_{0} $$and $$g(x)=b_{m}x^{m}+b_{m-1}x^{m-1}+cdots+b_{1}x+b_{0}.$$

    Then the value of $$lim_{xrightarrowinfty}frac{f(x)}{g(x)}$$ can be determined as follows:



    $bullet$ If $n>m$ and the sign of $a_n$ is the same as that of $b_m$ (i.e. $frac{a_{n}}{b_{m}}$ is positive) then the limit is $infty$



    $bullet$ If $n>m$ and the sign of $a_n$ is different from that of $b_m$, (i.e. $frac{a_{n}}{b_{m}}$ is negative) then the limit is $-infty$



    $bullet$ If $n<m$ then the limit is $0$.



    $bullet$ If $n=m$ then the limit is the value of $frac{a_{n}}{b_{m}}$.



    Note that this method is only for limits to infinity






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The post edit answer is correct. However, it can be obtained with little (or practically no) calculations at all by using the properties of end behaviour of polynomials.
      The end behaviour of a polynomial function can be defined, informally, as the behaviour of the function's graph as it approaches $pm infty$.



      For polynomial functions, the degree and the leading coefficient determine the end behaviour of the function. That means, as $x$ tends to infinity it is enough to analyse the leading term and coefficient to have an idea of what the graph will look like.



      Limits are, on one hand, asking about the end behaviour. In your case, you have two polynomial functions that make up a rational function. Therefore we can combine the property of end behaviour of polynomials and the quotient rule of limits, $${limlimits_{x to a} frac{{fleft( x right)}}{{gleft( x right)}} }={ frac{{limlimits_{x to a} fleft( x right)}}{{limlimits_{x to a} gleft( x right)}},;;;}kern-0.3pt
      {text{if};;limlimits_{x to a} gleft( x right) ne 0}$$       to quickly get the answer by considering $$lim_{xto infty} frac{x^3}{x^5} = lim_{x to infty} frac{1}{x^2} = frac{1}{infty} = 0$$ Of course, all this can be done in the head in just a matter of seconds.



      To Sum It Up



      Suppose you are given two rational functions $$f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+cdots+a_{1}x+a_{0} $$and $$g(x)=b_{m}x^{m}+b_{m-1}x^{m-1}+cdots+b_{1}x+b_{0}.$$

      Then the value of $$lim_{xrightarrowinfty}frac{f(x)}{g(x)}$$ can be determined as follows:



      $bullet$ If $n>m$ and the sign of $a_n$ is the same as that of $b_m$ (i.e. $frac{a_{n}}{b_{m}}$ is positive) then the limit is $infty$



      $bullet$ If $n>m$ and the sign of $a_n$ is different from that of $b_m$, (i.e. $frac{a_{n}}{b_{m}}$ is negative) then the limit is $-infty$



      $bullet$ If $n<m$ then the limit is $0$.



      $bullet$ If $n=m$ then the limit is the value of $frac{a_{n}}{b_{m}}$.



      Note that this method is only for limits to infinity






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The post edit answer is correct. However, it can be obtained with little (or practically no) calculations at all by using the properties of end behaviour of polynomials.
        The end behaviour of a polynomial function can be defined, informally, as the behaviour of the function's graph as it approaches $pm infty$.



        For polynomial functions, the degree and the leading coefficient determine the end behaviour of the function. That means, as $x$ tends to infinity it is enough to analyse the leading term and coefficient to have an idea of what the graph will look like.



        Limits are, on one hand, asking about the end behaviour. In your case, you have two polynomial functions that make up a rational function. Therefore we can combine the property of end behaviour of polynomials and the quotient rule of limits, $${limlimits_{x to a} frac{{fleft( x right)}}{{gleft( x right)}} }={ frac{{limlimits_{x to a} fleft( x right)}}{{limlimits_{x to a} gleft( x right)}},;;;}kern-0.3pt
        {text{if};;limlimits_{x to a} gleft( x right) ne 0}$$         to quickly get the answer by considering $$lim_{xto infty} frac{x^3}{x^5} = lim_{x to infty} frac{1}{x^2} = frac{1}{infty} = 0$$ Of course, all this can be done in the head in just a matter of seconds.



        To Sum It Up



        Suppose you are given two rational functions $$f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+cdots+a_{1}x+a_{0} $$and $$g(x)=b_{m}x^{m}+b_{m-1}x^{m-1}+cdots+b_{1}x+b_{0}.$$

        Then the value of $$lim_{xrightarrowinfty}frac{f(x)}{g(x)}$$ can be determined as follows:



        $bullet$ If $n>m$ and the sign of $a_n$ is the same as that of $b_m$ (i.e. $frac{a_{n}}{b_{m}}$ is positive) then the limit is $infty$



        $bullet$ If $n>m$ and the sign of $a_n$ is different from that of $b_m$, (i.e. $frac{a_{n}}{b_{m}}$ is negative) then the limit is $-infty$



        $bullet$ If $n<m$ then the limit is $0$.



        $bullet$ If $n=m$ then the limit is the value of $frac{a_{n}}{b_{m}}$.



        Note that this method is only for limits to infinity






        share|cite|improve this answer









        $endgroup$



        The post edit answer is correct. However, it can be obtained with little (or practically no) calculations at all by using the properties of end behaviour of polynomials.
        The end behaviour of a polynomial function can be defined, informally, as the behaviour of the function's graph as it approaches $pm infty$.



        For polynomial functions, the degree and the leading coefficient determine the end behaviour of the function. That means, as $x$ tends to infinity it is enough to analyse the leading term and coefficient to have an idea of what the graph will look like.



        Limits are, on one hand, asking about the end behaviour. In your case, you have two polynomial functions that make up a rational function. Therefore we can combine the property of end behaviour of polynomials and the quotient rule of limits, $${limlimits_{x to a} frac{{fleft( x right)}}{{gleft( x right)}} }={ frac{{limlimits_{x to a} fleft( x right)}}{{limlimits_{x to a} gleft( x right)}},;;;}kern-0.3pt
        {text{if};;limlimits_{x to a} gleft( x right) ne 0}$$         to quickly get the answer by considering $$lim_{xto infty} frac{x^3}{x^5} = lim_{x to infty} frac{1}{x^2} = frac{1}{infty} = 0$$ Of course, all this can be done in the head in just a matter of seconds.



        To Sum It Up



        Suppose you are given two rational functions $$f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+cdots+a_{1}x+a_{0} $$and $$g(x)=b_{m}x^{m}+b_{m-1}x^{m-1}+cdots+b_{1}x+b_{0}.$$

        Then the value of $$lim_{xrightarrowinfty}frac{f(x)}{g(x)}$$ can be determined as follows:



        $bullet$ If $n>m$ and the sign of $a_n$ is the same as that of $b_m$ (i.e. $frac{a_{n}}{b_{m}}$ is positive) then the limit is $infty$



        $bullet$ If $n>m$ and the sign of $a_n$ is different from that of $b_m$, (i.e. $frac{a_{n}}{b_{m}}$ is negative) then the limit is $-infty$



        $bullet$ If $n<m$ then the limit is $0$.



        $bullet$ If $n=m$ then the limit is the value of $frac{a_{n}}{b_{m}}$.



        Note that this method is only for limits to infinity







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 8 '18 at 11:45









        E.NoleE.Nole

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