Solve limit $lim_{xto1}frac{x^2-1}{ln(x)}$ without using L'Hôpital's rule?
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enter image description here
How to solve this limit?
$$lim_{xto1}frac{x^2-1}{ln(x)}=frac00$$
calculus limits limits-without-lhopital
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add a comment |
$begingroup$
enter image description here
How to solve this limit?
$$lim_{xto1}frac{x^2-1}{ln(x)}=frac00$$
calculus limits limits-without-lhopital
$endgroup$
$begingroup$
Welcome to the site! Please see the MathJax guide to see how to format your math.
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– Simply Beautiful Art
Jan 14 '17 at 14:48
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Not about the question, but that was a troll edit.
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– Cehhiro
Jan 14 '17 at 14:49
$begingroup$
Also, please do not mess up the pretty $LaTeX$...
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 14:50
add a comment |
$begingroup$
enter image description here
How to solve this limit?
$$lim_{xto1}frac{x^2-1}{ln(x)}=frac00$$
calculus limits limits-without-lhopital
$endgroup$
enter image description here
How to solve this limit?
$$lim_{xto1}frac{x^2-1}{ln(x)}=frac00$$
calculus limits limits-without-lhopital
calculus limits limits-without-lhopital
edited Jan 2 at 11:44
Nosrati
26.6k62354
26.6k62354
asked Jan 14 '17 at 14:47
Макс МотовиловМакс Мотовилов
92
92
$begingroup$
Welcome to the site! Please see the MathJax guide to see how to format your math.
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 14:48
$begingroup$
Not about the question, but that was a troll edit.
$endgroup$
– Cehhiro
Jan 14 '17 at 14:49
$begingroup$
Also, please do not mess up the pretty $LaTeX$...
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 14:50
add a comment |
$begingroup$
Welcome to the site! Please see the MathJax guide to see how to format your math.
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 14:48
$begingroup$
Not about the question, but that was a troll edit.
$endgroup$
– Cehhiro
Jan 14 '17 at 14:49
$begingroup$
Also, please do not mess up the pretty $LaTeX$...
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 14:50
$begingroup$
Welcome to the site! Please see the MathJax guide to see how to format your math.
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 14:48
$begingroup$
Welcome to the site! Please see the MathJax guide to see how to format your math.
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 14:48
$begingroup$
Not about the question, but that was a troll edit.
$endgroup$
– Cehhiro
Jan 14 '17 at 14:49
$begingroup$
Not about the question, but that was a troll edit.
$endgroup$
– Cehhiro
Jan 14 '17 at 14:49
$begingroup$
Also, please do not mess up the pretty $LaTeX$...
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 14:50
$begingroup$
Also, please do not mess up the pretty $LaTeX$...
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 14:50
add a comment |
5 Answers
5
active
oldest
votes
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Set $x-1=hiff x=1+h$
$$lim_{xto1}dfrac{x^2-1}{ln x}=dfrac{lim_{hto0}dfrac{(h+1)^2-1}h}{lim_{hto0}dfrac{ln(1+h)}h}=?$$
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$begingroup$
:D That was pretty neat to see.
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– Simply Beautiful Art
Jan 14 '17 at 14:53
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Very neat, but doesn't he still need L'Hôpital for the denominator?
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– Cehhiro
Jan 14 '17 at 14:54
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@O.VonSeckendorff You need something to get past the $ln$, whether it be known limits or series expansion.
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 14:56
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@O.VonSeckendorff, This is an elementary formula of limit, readily available from $$lim_{xto0}dfrac{e^x-1}x=1$$ See math.stackexchange.com/questions/690898/…
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– lab bhattacharjee
Jan 14 '17 at 14:56
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@labbhattacharjee I didn't know/remember that one. Good to know.
$endgroup$
– Cehhiro
Jan 14 '17 at 14:57
|
show 5 more comments
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In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities
$$bbox[5px,border:2px solid #C0A000]{frac{x-1}{x}le log(x)le x-1} tag 1$$
for $x>0$.
Hence, we have from $(1)$
$$frac{x^2-1}{x-1}le frac{x^2-1}{log(x)}le frac{x(x^2-1)}{x-1} tag 2$$
whence application of the squeeze theorem yields the coveted limit
$$lim_{xto 1}frac{x^2-1}{log(x)}=2$$
$endgroup$
1
$begingroup$
Plz. Check (1).
$endgroup$
– Nosrati
Jan 14 '17 at 15:00
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@MyGlasses Thank you; I've edited. -Mark
$endgroup$
– Mark Viola
Jan 14 '17 at 15:04
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Haha, nice... :-D
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 15:09
$begingroup$
@SimpleArt Thank you. And if you follow the link, it will show a way forward that begins with the limit definition of the exponential function.
$endgroup$
– Mark Viola
Jan 14 '17 at 15:11
add a comment |
$begingroup$
With substitution $x=e^t$ we have:
$$lim_{xto1}frac{x^2-1}{ln x}=lim_{tto0}frac{e^{2t}-1}{t}=lim_{tto0}frac{1+2t+frac{(2t)^2}{2}-1}{t}=2$$
where
$$e^theta=1+theta+frac{theta^2}{2}$$
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We can employ $$lim_{xto0}dfrac{e^x-1}x=1$$
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– lab bhattacharjee
Jan 14 '17 at 15:00
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@labbhattacharjee from where? without proof?
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– Nosrati
Jan 14 '17 at 15:01
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Which identities can be used for limit ? math.stackexchange.com/questions/42679/… OR 2000clicks.com/MathHelp/CalculusLimitExponential.aspx
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– lab bhattacharjee
Jan 14 '17 at 15:04
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I mean, it is limit definition to derivative of $e^x$ at $x=0$.
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 15:09
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@labbhattacharjee For using this limit, according to your links, all apply l'hopital rule, derivative definition and Taylor expansion. My suppose is the part of Taylor expansion also, but I don't know where can find a proof without these techniques.
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– Nosrati
Jan 14 '17 at 15:09
add a comment |
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Notice that
$$ln(x)=(x-1)-frac12(x-1)^2+mathcal O(x^3)$$
and
$$x^2-1=(x-1)(x+1)$$
So the reciprocal of the limit is
$$frac1L=lim_{xto1}frac1{x+1}-frac{(x-1)}{2(x+1)}+mathcal Oleft(frac{(x-1)^2}{x+1}right)=frac12$$
So that
$$L=2$$
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add a comment |
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By a shift of the variable your limit is equivalent to
$$lim_{tto0}frac{t(t+1)}{ln(1+t)}=frac{lim_{tto0}t+1}{lim_{tto0}dfrac{ln(1+t)}t}=2.$$
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add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Set $x-1=hiff x=1+h$
$$lim_{xto1}dfrac{x^2-1}{ln x}=dfrac{lim_{hto0}dfrac{(h+1)^2-1}h}{lim_{hto0}dfrac{ln(1+h)}h}=?$$
$endgroup$
$begingroup$
:D That was pretty neat to see.
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 14:53
$begingroup$
Very neat, but doesn't he still need L'Hôpital for the denominator?
$endgroup$
– Cehhiro
Jan 14 '17 at 14:54
$begingroup$
@O.VonSeckendorff You need something to get past the $ln$, whether it be known limits or series expansion.
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 14:56
$begingroup$
@O.VonSeckendorff, This is an elementary formula of limit, readily available from $$lim_{xto0}dfrac{e^x-1}x=1$$ See math.stackexchange.com/questions/690898/…
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– lab bhattacharjee
Jan 14 '17 at 14:56
$begingroup$
@labbhattacharjee I didn't know/remember that one. Good to know.
$endgroup$
– Cehhiro
Jan 14 '17 at 14:57
|
show 5 more comments
$begingroup$
Set $x-1=hiff x=1+h$
$$lim_{xto1}dfrac{x^2-1}{ln x}=dfrac{lim_{hto0}dfrac{(h+1)^2-1}h}{lim_{hto0}dfrac{ln(1+h)}h}=?$$
$endgroup$
$begingroup$
:D That was pretty neat to see.
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 14:53
$begingroup$
Very neat, but doesn't he still need L'Hôpital for the denominator?
$endgroup$
– Cehhiro
Jan 14 '17 at 14:54
$begingroup$
@O.VonSeckendorff You need something to get past the $ln$, whether it be known limits or series expansion.
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 14:56
$begingroup$
@O.VonSeckendorff, This is an elementary formula of limit, readily available from $$lim_{xto0}dfrac{e^x-1}x=1$$ See math.stackexchange.com/questions/690898/…
$endgroup$
– lab bhattacharjee
Jan 14 '17 at 14:56
$begingroup$
@labbhattacharjee I didn't know/remember that one. Good to know.
$endgroup$
– Cehhiro
Jan 14 '17 at 14:57
|
show 5 more comments
$begingroup$
Set $x-1=hiff x=1+h$
$$lim_{xto1}dfrac{x^2-1}{ln x}=dfrac{lim_{hto0}dfrac{(h+1)^2-1}h}{lim_{hto0}dfrac{ln(1+h)}h}=?$$
$endgroup$
Set $x-1=hiff x=1+h$
$$lim_{xto1}dfrac{x^2-1}{ln x}=dfrac{lim_{hto0}dfrac{(h+1)^2-1}h}{lim_{hto0}dfrac{ln(1+h)}h}=?$$
edited Jan 14 '17 at 14:53
answered Jan 14 '17 at 14:52
lab bhattacharjeelab bhattacharjee
226k15157275
226k15157275
$begingroup$
:D That was pretty neat to see.
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 14:53
$begingroup$
Very neat, but doesn't he still need L'Hôpital for the denominator?
$endgroup$
– Cehhiro
Jan 14 '17 at 14:54
$begingroup$
@O.VonSeckendorff You need something to get past the $ln$, whether it be known limits or series expansion.
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 14:56
$begingroup$
@O.VonSeckendorff, This is an elementary formula of limit, readily available from $$lim_{xto0}dfrac{e^x-1}x=1$$ See math.stackexchange.com/questions/690898/…
$endgroup$
– lab bhattacharjee
Jan 14 '17 at 14:56
$begingroup$
@labbhattacharjee I didn't know/remember that one. Good to know.
$endgroup$
– Cehhiro
Jan 14 '17 at 14:57
|
show 5 more comments
$begingroup$
:D That was pretty neat to see.
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 14:53
$begingroup$
Very neat, but doesn't he still need L'Hôpital for the denominator?
$endgroup$
– Cehhiro
Jan 14 '17 at 14:54
$begingroup$
@O.VonSeckendorff You need something to get past the $ln$, whether it be known limits or series expansion.
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 14:56
$begingroup$
@O.VonSeckendorff, This is an elementary formula of limit, readily available from $$lim_{xto0}dfrac{e^x-1}x=1$$ See math.stackexchange.com/questions/690898/…
$endgroup$
– lab bhattacharjee
Jan 14 '17 at 14:56
$begingroup$
@labbhattacharjee I didn't know/remember that one. Good to know.
$endgroup$
– Cehhiro
Jan 14 '17 at 14:57
$begingroup$
:D That was pretty neat to see.
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 14:53
$begingroup$
:D That was pretty neat to see.
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 14:53
$begingroup$
Very neat, but doesn't he still need L'Hôpital for the denominator?
$endgroup$
– Cehhiro
Jan 14 '17 at 14:54
$begingroup$
Very neat, but doesn't he still need L'Hôpital for the denominator?
$endgroup$
– Cehhiro
Jan 14 '17 at 14:54
$begingroup$
@O.VonSeckendorff You need something to get past the $ln$, whether it be known limits or series expansion.
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 14:56
$begingroup$
@O.VonSeckendorff You need something to get past the $ln$, whether it be known limits or series expansion.
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 14:56
$begingroup$
@O.VonSeckendorff, This is an elementary formula of limit, readily available from $$lim_{xto0}dfrac{e^x-1}x=1$$ See math.stackexchange.com/questions/690898/…
$endgroup$
– lab bhattacharjee
Jan 14 '17 at 14:56
$begingroup$
@O.VonSeckendorff, This is an elementary formula of limit, readily available from $$lim_{xto0}dfrac{e^x-1}x=1$$ See math.stackexchange.com/questions/690898/…
$endgroup$
– lab bhattacharjee
Jan 14 '17 at 14:56
$begingroup$
@labbhattacharjee I didn't know/remember that one. Good to know.
$endgroup$
– Cehhiro
Jan 14 '17 at 14:57
$begingroup$
@labbhattacharjee I didn't know/remember that one. Good to know.
$endgroup$
– Cehhiro
Jan 14 '17 at 14:57
|
show 5 more comments
$begingroup$
In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities
$$bbox[5px,border:2px solid #C0A000]{frac{x-1}{x}le log(x)le x-1} tag 1$$
for $x>0$.
Hence, we have from $(1)$
$$frac{x^2-1}{x-1}le frac{x^2-1}{log(x)}le frac{x(x^2-1)}{x-1} tag 2$$
whence application of the squeeze theorem yields the coveted limit
$$lim_{xto 1}frac{x^2-1}{log(x)}=2$$
$endgroup$
1
$begingroup$
Plz. Check (1).
$endgroup$
– Nosrati
Jan 14 '17 at 15:00
$begingroup$
@MyGlasses Thank you; I've edited. -Mark
$endgroup$
– Mark Viola
Jan 14 '17 at 15:04
$begingroup$
Haha, nice... :-D
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 15:09
$begingroup$
@SimpleArt Thank you. And if you follow the link, it will show a way forward that begins with the limit definition of the exponential function.
$endgroup$
– Mark Viola
Jan 14 '17 at 15:11
add a comment |
$begingroup$
In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities
$$bbox[5px,border:2px solid #C0A000]{frac{x-1}{x}le log(x)le x-1} tag 1$$
for $x>0$.
Hence, we have from $(1)$
$$frac{x^2-1}{x-1}le frac{x^2-1}{log(x)}le frac{x(x^2-1)}{x-1} tag 2$$
whence application of the squeeze theorem yields the coveted limit
$$lim_{xto 1}frac{x^2-1}{log(x)}=2$$
$endgroup$
1
$begingroup$
Plz. Check (1).
$endgroup$
– Nosrati
Jan 14 '17 at 15:00
$begingroup$
@MyGlasses Thank you; I've edited. -Mark
$endgroup$
– Mark Viola
Jan 14 '17 at 15:04
$begingroup$
Haha, nice... :-D
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 15:09
$begingroup$
@SimpleArt Thank you. And if you follow the link, it will show a way forward that begins with the limit definition of the exponential function.
$endgroup$
– Mark Viola
Jan 14 '17 at 15:11
add a comment |
$begingroup$
In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities
$$bbox[5px,border:2px solid #C0A000]{frac{x-1}{x}le log(x)le x-1} tag 1$$
for $x>0$.
Hence, we have from $(1)$
$$frac{x^2-1}{x-1}le frac{x^2-1}{log(x)}le frac{x(x^2-1)}{x-1} tag 2$$
whence application of the squeeze theorem yields the coveted limit
$$lim_{xto 1}frac{x^2-1}{log(x)}=2$$
$endgroup$
In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities
$$bbox[5px,border:2px solid #C0A000]{frac{x-1}{x}le log(x)le x-1} tag 1$$
for $x>0$.
Hence, we have from $(1)$
$$frac{x^2-1}{x-1}le frac{x^2-1}{log(x)}le frac{x(x^2-1)}{x-1} tag 2$$
whence application of the squeeze theorem yields the coveted limit
$$lim_{xto 1}frac{x^2-1}{log(x)}=2$$
edited Apr 13 '17 at 12:21
Community♦
1
1
answered Jan 14 '17 at 14:58
Mark ViolaMark Viola
133k1277174
133k1277174
1
$begingroup$
Plz. Check (1).
$endgroup$
– Nosrati
Jan 14 '17 at 15:00
$begingroup$
@MyGlasses Thank you; I've edited. -Mark
$endgroup$
– Mark Viola
Jan 14 '17 at 15:04
$begingroup$
Haha, nice... :-D
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 15:09
$begingroup$
@SimpleArt Thank you. And if you follow the link, it will show a way forward that begins with the limit definition of the exponential function.
$endgroup$
– Mark Viola
Jan 14 '17 at 15:11
add a comment |
1
$begingroup$
Plz. Check (1).
$endgroup$
– Nosrati
Jan 14 '17 at 15:00
$begingroup$
@MyGlasses Thank you; I've edited. -Mark
$endgroup$
– Mark Viola
Jan 14 '17 at 15:04
$begingroup$
Haha, nice... :-D
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 15:09
$begingroup$
@SimpleArt Thank you. And if you follow the link, it will show a way forward that begins with the limit definition of the exponential function.
$endgroup$
– Mark Viola
Jan 14 '17 at 15:11
1
1
$begingroup$
Plz. Check (1).
$endgroup$
– Nosrati
Jan 14 '17 at 15:00
$begingroup$
Plz. Check (1).
$endgroup$
– Nosrati
Jan 14 '17 at 15:00
$begingroup$
@MyGlasses Thank you; I've edited. -Mark
$endgroup$
– Mark Viola
Jan 14 '17 at 15:04
$begingroup$
@MyGlasses Thank you; I've edited. -Mark
$endgroup$
– Mark Viola
Jan 14 '17 at 15:04
$begingroup$
Haha, nice... :-D
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 15:09
$begingroup$
Haha, nice... :-D
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 15:09
$begingroup$
@SimpleArt Thank you. And if you follow the link, it will show a way forward that begins with the limit definition of the exponential function.
$endgroup$
– Mark Viola
Jan 14 '17 at 15:11
$begingroup$
@SimpleArt Thank you. And if you follow the link, it will show a way forward that begins with the limit definition of the exponential function.
$endgroup$
– Mark Viola
Jan 14 '17 at 15:11
add a comment |
$begingroup$
With substitution $x=e^t$ we have:
$$lim_{xto1}frac{x^2-1}{ln x}=lim_{tto0}frac{e^{2t}-1}{t}=lim_{tto0}frac{1+2t+frac{(2t)^2}{2}-1}{t}=2$$
where
$$e^theta=1+theta+frac{theta^2}{2}$$
$endgroup$
$begingroup$
We can employ $$lim_{xto0}dfrac{e^x-1}x=1$$
$endgroup$
– lab bhattacharjee
Jan 14 '17 at 15:00
$begingroup$
@labbhattacharjee from where? without proof?
$endgroup$
– Nosrati
Jan 14 '17 at 15:01
$begingroup$
Which identities can be used for limit ? math.stackexchange.com/questions/42679/… OR 2000clicks.com/MathHelp/CalculusLimitExponential.aspx
$endgroup$
– lab bhattacharjee
Jan 14 '17 at 15:04
$begingroup$
I mean, it is limit definition to derivative of $e^x$ at $x=0$.
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 15:09
$begingroup$
@labbhattacharjee For using this limit, according to your links, all apply l'hopital rule, derivative definition and Taylor expansion. My suppose is the part of Taylor expansion also, but I don't know where can find a proof without these techniques.
$endgroup$
– Nosrati
Jan 14 '17 at 15:09
add a comment |
$begingroup$
With substitution $x=e^t$ we have:
$$lim_{xto1}frac{x^2-1}{ln x}=lim_{tto0}frac{e^{2t}-1}{t}=lim_{tto0}frac{1+2t+frac{(2t)^2}{2}-1}{t}=2$$
where
$$e^theta=1+theta+frac{theta^2}{2}$$
$endgroup$
$begingroup$
We can employ $$lim_{xto0}dfrac{e^x-1}x=1$$
$endgroup$
– lab bhattacharjee
Jan 14 '17 at 15:00
$begingroup$
@labbhattacharjee from where? without proof?
$endgroup$
– Nosrati
Jan 14 '17 at 15:01
$begingroup$
Which identities can be used for limit ? math.stackexchange.com/questions/42679/… OR 2000clicks.com/MathHelp/CalculusLimitExponential.aspx
$endgroup$
– lab bhattacharjee
Jan 14 '17 at 15:04
$begingroup$
I mean, it is limit definition to derivative of $e^x$ at $x=0$.
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– Simply Beautiful Art
Jan 14 '17 at 15:09
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@labbhattacharjee For using this limit, according to your links, all apply l'hopital rule, derivative definition and Taylor expansion. My suppose is the part of Taylor expansion also, but I don't know where can find a proof without these techniques.
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– Nosrati
Jan 14 '17 at 15:09
add a comment |
$begingroup$
With substitution $x=e^t$ we have:
$$lim_{xto1}frac{x^2-1}{ln x}=lim_{tto0}frac{e^{2t}-1}{t}=lim_{tto0}frac{1+2t+frac{(2t)^2}{2}-1}{t}=2$$
where
$$e^theta=1+theta+frac{theta^2}{2}$$
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With substitution $x=e^t$ we have:
$$lim_{xto1}frac{x^2-1}{ln x}=lim_{tto0}frac{e^{2t}-1}{t}=lim_{tto0}frac{1+2t+frac{(2t)^2}{2}-1}{t}=2$$
where
$$e^theta=1+theta+frac{theta^2}{2}$$
edited Jan 14 '17 at 15:04
answered Jan 14 '17 at 14:58
NosratiNosrati
26.6k62354
26.6k62354
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We can employ $$lim_{xto0}dfrac{e^x-1}x=1$$
$endgroup$
– lab bhattacharjee
Jan 14 '17 at 15:00
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@labbhattacharjee from where? without proof?
$endgroup$
– Nosrati
Jan 14 '17 at 15:01
$begingroup$
Which identities can be used for limit ? math.stackexchange.com/questions/42679/… OR 2000clicks.com/MathHelp/CalculusLimitExponential.aspx
$endgroup$
– lab bhattacharjee
Jan 14 '17 at 15:04
$begingroup$
I mean, it is limit definition to derivative of $e^x$ at $x=0$.
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 15:09
$begingroup$
@labbhattacharjee For using this limit, according to your links, all apply l'hopital rule, derivative definition and Taylor expansion. My suppose is the part of Taylor expansion also, but I don't know where can find a proof without these techniques.
$endgroup$
– Nosrati
Jan 14 '17 at 15:09
add a comment |
$begingroup$
We can employ $$lim_{xto0}dfrac{e^x-1}x=1$$
$endgroup$
– lab bhattacharjee
Jan 14 '17 at 15:00
$begingroup$
@labbhattacharjee from where? without proof?
$endgroup$
– Nosrati
Jan 14 '17 at 15:01
$begingroup$
Which identities can be used for limit ? math.stackexchange.com/questions/42679/… OR 2000clicks.com/MathHelp/CalculusLimitExponential.aspx
$endgroup$
– lab bhattacharjee
Jan 14 '17 at 15:04
$begingroup$
I mean, it is limit definition to derivative of $e^x$ at $x=0$.
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 15:09
$begingroup$
@labbhattacharjee For using this limit, according to your links, all apply l'hopital rule, derivative definition and Taylor expansion. My suppose is the part of Taylor expansion also, but I don't know where can find a proof without these techniques.
$endgroup$
– Nosrati
Jan 14 '17 at 15:09
$begingroup$
We can employ $$lim_{xto0}dfrac{e^x-1}x=1$$
$endgroup$
– lab bhattacharjee
Jan 14 '17 at 15:00
$begingroup$
We can employ $$lim_{xto0}dfrac{e^x-1}x=1$$
$endgroup$
– lab bhattacharjee
Jan 14 '17 at 15:00
$begingroup$
@labbhattacharjee from where? without proof?
$endgroup$
– Nosrati
Jan 14 '17 at 15:01
$begingroup$
@labbhattacharjee from where? without proof?
$endgroup$
– Nosrati
Jan 14 '17 at 15:01
$begingroup$
Which identities can be used for limit ? math.stackexchange.com/questions/42679/… OR 2000clicks.com/MathHelp/CalculusLimitExponential.aspx
$endgroup$
– lab bhattacharjee
Jan 14 '17 at 15:04
$begingroup$
Which identities can be used for limit ? math.stackexchange.com/questions/42679/… OR 2000clicks.com/MathHelp/CalculusLimitExponential.aspx
$endgroup$
– lab bhattacharjee
Jan 14 '17 at 15:04
$begingroup$
I mean, it is limit definition to derivative of $e^x$ at $x=0$.
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 15:09
$begingroup$
I mean, it is limit definition to derivative of $e^x$ at $x=0$.
$endgroup$
– Simply Beautiful Art
Jan 14 '17 at 15:09
$begingroup$
@labbhattacharjee For using this limit, according to your links, all apply l'hopital rule, derivative definition and Taylor expansion. My suppose is the part of Taylor expansion also, but I don't know where can find a proof without these techniques.
$endgroup$
– Nosrati
Jan 14 '17 at 15:09
$begingroup$
@labbhattacharjee For using this limit, according to your links, all apply l'hopital rule, derivative definition and Taylor expansion. My suppose is the part of Taylor expansion also, but I don't know where can find a proof without these techniques.
$endgroup$
– Nosrati
Jan 14 '17 at 15:09
add a comment |
$begingroup$
Notice that
$$ln(x)=(x-1)-frac12(x-1)^2+mathcal O(x^3)$$
and
$$x^2-1=(x-1)(x+1)$$
So the reciprocal of the limit is
$$frac1L=lim_{xto1}frac1{x+1}-frac{(x-1)}{2(x+1)}+mathcal Oleft(frac{(x-1)^2}{x+1}right)=frac12$$
So that
$$L=2$$
$endgroup$
add a comment |
$begingroup$
Notice that
$$ln(x)=(x-1)-frac12(x-1)^2+mathcal O(x^3)$$
and
$$x^2-1=(x-1)(x+1)$$
So the reciprocal of the limit is
$$frac1L=lim_{xto1}frac1{x+1}-frac{(x-1)}{2(x+1)}+mathcal Oleft(frac{(x-1)^2}{x+1}right)=frac12$$
So that
$$L=2$$
$endgroup$
add a comment |
$begingroup$
Notice that
$$ln(x)=(x-1)-frac12(x-1)^2+mathcal O(x^3)$$
and
$$x^2-1=(x-1)(x+1)$$
So the reciprocal of the limit is
$$frac1L=lim_{xto1}frac1{x+1}-frac{(x-1)}{2(x+1)}+mathcal Oleft(frac{(x-1)^2}{x+1}right)=frac12$$
So that
$$L=2$$
$endgroup$
Notice that
$$ln(x)=(x-1)-frac12(x-1)^2+mathcal O(x^3)$$
and
$$x^2-1=(x-1)(x+1)$$
So the reciprocal of the limit is
$$frac1L=lim_{xto1}frac1{x+1}-frac{(x-1)}{2(x+1)}+mathcal Oleft(frac{(x-1)^2}{x+1}right)=frac12$$
So that
$$L=2$$
answered Jan 14 '17 at 14:52
Simply Beautiful ArtSimply Beautiful Art
50.7k579183
50.7k579183
add a comment |
add a comment |
$begingroup$
By a shift of the variable your limit is equivalent to
$$lim_{tto0}frac{t(t+1)}{ln(1+t)}=frac{lim_{tto0}t+1}{lim_{tto0}dfrac{ln(1+t)}t}=2.$$
$endgroup$
add a comment |
$begingroup$
By a shift of the variable your limit is equivalent to
$$lim_{tto0}frac{t(t+1)}{ln(1+t)}=frac{lim_{tto0}t+1}{lim_{tto0}dfrac{ln(1+t)}t}=2.$$
$endgroup$
add a comment |
$begingroup$
By a shift of the variable your limit is equivalent to
$$lim_{tto0}frac{t(t+1)}{ln(1+t)}=frac{lim_{tto0}t+1}{lim_{tto0}dfrac{ln(1+t)}t}=2.$$
$endgroup$
By a shift of the variable your limit is equivalent to
$$lim_{tto0}frac{t(t+1)}{ln(1+t)}=frac{lim_{tto0}t+1}{lim_{tto0}dfrac{ln(1+t)}t}=2.$$
answered Jan 14 '17 at 15:32
Yves DaoustYves Daoust
129k675227
129k675227
add a comment |
add a comment |
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– Simply Beautiful Art
Jan 14 '17 at 14:48
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Not about the question, but that was a troll edit.
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– Cehhiro
Jan 14 '17 at 14:49
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Also, please do not mess up the pretty $LaTeX$...
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– Simply Beautiful Art
Jan 14 '17 at 14:50