How to find the derivative of the function at $0$
How to find the derivative of the function at $0$
$f(x)=e^{-frac{1}{x^2}}sinfrac{1}{x}$ if $xne0$
$=0$ if $x=0$
I cannnot evaluate the limit $$lim_{xto0}frac{e^{-frac{1}{x^2}}sinfrac{1}{x}}{x}.$$
L'Hospital is not working here.
Please help.
real-analysis
asked Nov 22 at 3:11
Jave
458114
add a comment |
How to find the derivative of the function at $0$
$f(x)=e^{-frac{1}{x^2}}sinfrac{1}{x}$ if $xne0$
$=0$ if $x=0$
I cannnot evaluate the limit $$lim_{xto0}frac{e^{-frac{1}{x^2}}sinfrac{1}{x}}{x}.$$
L'Hospital is not working here.
Please help.
real-analysis
asked Nov 22 at 3:11
Jave
458114
add a comment |
How to find the derivative of the function at $0$
$f(x)=e^{-frac{1}{x^2}}sinfrac{1}{x}$ if $xne0$
$=0$ if $x=0$
I cannnot evaluate the limit $$lim_{xto0}frac{e^{-frac{1}{x^2}}sinfrac{1}{x}}{x}.$$
L'Hospital is not working here.
Please help.
real-analysis
asked Nov 22 at 3:11
Jave
458114
How to find the derivative of the function at $0$
$f(x)=e^{-frac{1}{x^2}}sinfrac{1}{x}$ if $xne0$
$=0$ if $x=0$
I cannnot evaluate the limit $$lim_{xto0}frac{e^{-frac{1}{x^2}}sinfrac{1}{x}}{x}.$$
L'Hospital is not working here.
Please help.
real-analysis
real-analysis
asked Nov 22 at 3:11
Jave
458114
asked Nov 22 at 3:11
Jave
458114
asked Nov 22 at 3:11
Jave
458114
asked Nov 22 at 3:11
Jave
458114
asked Nov 22 at 3:11
Jave
458114
458114
add a comment |
add a comment |
3 Answers
3
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oldest
votes
Hint 1: $-1le sin(1/x)le 1$.
Hint 2: $displaystyle lim_{xto 0}frac{1}{x}e^{-1/x^2} = lim_{yto pminfty} ycdot e^{-y^2}$.
answered Nov 22 at 3:23
Torsten Schoeneberg
3,6812833
I have tried this but could not solve yet the rest of the limit.
– Jave
Nov 22 at 3:25
add a comment |
You already know that $|sin(1/x)|le 1$. This helps you only if the rest is $0$. So do a change of variable $y=1/x$ and use $e^{-frac 1{x^2}}=frac 1 {e^{y^2}}$. Now you need to calculate $$lim_{ytopminfty}frac y{e^{y^2}}$$
answered Nov 22 at 3:34
Andrei
10.8k21025
add a comment |
$-1le sin(1/x)le 1 ::text{and let $frac1x=t$}implies -dfrac{t}{e^{t^2}}leq sin tdfrac{t} {e^{t^2}} leq dfrac{t}{e^{t^2}}$ Taking limit as $trightarrow infty$ and using L'hopital rule we get $displaystyle lim_{trightarrowinfty}sin tdfrac{t} {e^{t^2}}=0$.
answered Nov 22 at 3:40
Yadati Kiran
1,698519
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Hint 1: $-1le sin(1/x)le 1$.
Hint 2: $displaystyle lim_{xto 0}frac{1}{x}e^{-1/x^2} = lim_{yto pminfty} ycdot e^{-y^2}$.
answered Nov 22 at 3:23
Torsten Schoeneberg
3,6812833
I have tried this but could not solve yet the rest of the limit.
– Jave
Nov 22 at 3:25
add a comment |
Hint 1: $-1le sin(1/x)le 1$.
Hint 2: $displaystyle lim_{xto 0}frac{1}{x}e^{-1/x^2} = lim_{yto pminfty} ycdot e^{-y^2}$.
answered Nov 22 at 3:23
Torsten Schoeneberg
3,6812833
I have tried this but could not solve yet the rest of the limit.
– Jave
Nov 22 at 3:25
add a comment |
Hint 1: $-1le sin(1/x)le 1$.
Hint 2: $displaystyle lim_{xto 0}frac{1}{x}e^{-1/x^2} = lim_{yto pminfty} ycdot e^{-y^2}$.
answered Nov 22 at 3:23
Torsten Schoeneberg
3,6812833
Hint 1: $-1le sin(1/x)le 1$.
Hint 2: $displaystyle lim_{xto 0}frac{1}{x}e^{-1/x^2} = lim_{yto pminfty} ycdot e^{-y^2}$.
answered Nov 22 at 3:23
Torsten Schoeneberg
3,6812833
edited Nov 22 at 3:36
answered Nov 22 at 3:23
Torsten Schoeneberg
3,6812833
answered Nov 22 at 3:23
Torsten Schoeneberg
3,6812833
answered Nov 22 at 3:23
Torsten Schoeneberg
3,6812833
3,6812833
I have tried this but could not solve yet the rest of the limit.
– Jave
Nov 22 at 3:25
add a comment |
I have tried this but could not solve yet the rest of the limit.
– Jave
Nov 22 at 3:25
I have tried this but could not solve yet the rest of the limit.
– Jave
Nov 22 at 3:25
I have tried this but could not solve yet the rest of the limit.
– Jave
Nov 22 at 3:25
add a comment |
You already know that $|sin(1/x)|le 1$. This helps you only if the rest is $0$. So do a change of variable $y=1/x$ and use $e^{-frac 1{x^2}}=frac 1 {e^{y^2}}$. Now you need to calculate $$lim_{ytopminfty}frac y{e^{y^2}}$$
answered Nov 22 at 3:34
Andrei
10.8k21025
add a comment |
You already know that $|sin(1/x)|le 1$. This helps you only if the rest is $0$. So do a change of variable $y=1/x$ and use $e^{-frac 1{x^2}}=frac 1 {e^{y^2}}$. Now you need to calculate $$lim_{ytopminfty}frac y{e^{y^2}}$$
answered Nov 22 at 3:34
Andrei
10.8k21025
add a comment |
You already know that $|sin(1/x)|le 1$. This helps you only if the rest is $0$. So do a change of variable $y=1/x$ and use $e^{-frac 1{x^2}}=frac 1 {e^{y^2}}$. Now you need to calculate $$lim_{ytopminfty}frac y{e^{y^2}}$$
answered Nov 22 at 3:34
Andrei
10.8k21025
You already know that $|sin(1/x)|le 1$. This helps you only if the rest is $0$. So do a change of variable $y=1/x$ and use $e^{-frac 1{x^2}}=frac 1 {e^{y^2}}$. Now you need to calculate $$lim_{ytopminfty}frac y{e^{y^2}}$$
answered Nov 22 at 3:34
Andrei
10.8k21025
answered Nov 22 at 3:34
Andrei
10.8k21025
answered Nov 22 at 3:34
Andrei
10.8k21025
answered Nov 22 at 3:34
Andrei
10.8k21025
10.8k21025
add a comment |
add a comment |
$-1le sin(1/x)le 1 ::text{and let $frac1x=t$}implies -dfrac{t}{e^{t^2}}leq sin tdfrac{t} {e^{t^2}} leq dfrac{t}{e^{t^2}}$ Taking limit as $trightarrow infty$ and using L'hopital rule we get $displaystyle lim_{trightarrowinfty}sin tdfrac{t} {e^{t^2}}=0$.
answered Nov 22 at 3:40
Yadati Kiran
1,698519
add a comment |
$-1le sin(1/x)le 1 ::text{and let $frac1x=t$}implies -dfrac{t}{e^{t^2}}leq sin tdfrac{t} {e^{t^2}} leq dfrac{t}{e^{t^2}}$ Taking limit as $trightarrow infty$ and using L'hopital rule we get $displaystyle lim_{trightarrowinfty}sin tdfrac{t} {e^{t^2}}=0$.
answered Nov 22 at 3:40
Yadati Kiran
1,698519
add a comment |
$-1le sin(1/x)le 1 ::text{and let $frac1x=t$}implies -dfrac{t}{e^{t^2}}leq sin tdfrac{t} {e^{t^2}} leq dfrac{t}{e^{t^2}}$ Taking limit as $trightarrow infty$ and using L'hopital rule we get $displaystyle lim_{trightarrowinfty}sin tdfrac{t} {e^{t^2}}=0$.
answered Nov 22 at 3:40
Yadati Kiran
1,698519
$-1le sin(1/x)le 1 ::text{and let $frac1x=t$}implies -dfrac{t}{e^{t^2}}leq sin tdfrac{t} {e^{t^2}} leq dfrac{t}{e^{t^2}}$ Taking limit as $trightarrow infty$ and using L'hopital rule we get $displaystyle lim_{trightarrowinfty}sin tdfrac{t} {e^{t^2}}=0$.
answered Nov 22 at 3:40
Yadati Kiran
1,698519
answered Nov 22 at 3:40
Yadati Kiran
1,698519
answered Nov 22 at 3:40
Yadati Kiran
1,698519
answered Nov 22 at 3:40
Yadati Kiran
1,698519
1,698519
add a comment |
add a comment |
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