Mixing
How to calculate this limit :$ lim_{xto 1} x^{(frac{1}{x-1})}$
$begingroup$
The graph of this function shows that the limit should be approximately $2.7$, but I don't know how to approach it mathematically, the form after direct putting of $x = 1$ is $1^{infty}$, which is an indeterminate form, so we can proceed with L'-H$hat{o}$pital's rule, but I don't really know how to do that. Please help me with this limit.
calculus limits
asked Jan 16 at 1:39
SimranSimran
153
$endgroup$
add a comment |
$begingroup$
The graph of this function shows that the limit should be approximately $2.7$, but I don't know how to approach it mathematically, the form after direct putting of $x = 1$ is $1^{infty}$, which is an indeterminate form, so we can proceed with L'-H$hat{o}$pital's rule, but I don't really know how to do that. Please help me with this limit.
calculus limits
asked Jan 16 at 1:39
SimranSimran
153
$endgroup$
$begingroup$
Note that L'Hopitals rule only applies to the $frac{0}{0}$ and $frac{infty}{infty}$ indeterminate forms.
$endgroup$
– H Huang
Jan 16 at 1:42
$begingroup$
Hint: $x = 1 + frac{1}{frac{1}{(x-1)}}$
$endgroup$
– user626177
Jan 16 at 1:43
add a comment |
$begingroup$
The graph of this function shows that the limit should be approximately $2.7$, but I don't know how to approach it mathematically, the form after direct putting of $x = 1$ is $1^{infty}$, which is an indeterminate form, so we can proceed with L'-H$hat{o}$pital's rule, but I don't really know how to do that. Please help me with this limit.
calculus limits
asked Jan 16 at 1:39
SimranSimran
153
$endgroup$
The graph of this function shows that the limit should be approximately $2.7$, but I don't know how to approach it mathematically, the form after direct putting of $x = 1$ is $1^{infty}$, which is an indeterminate form, so we can proceed with L'-H$hat{o}$pital's rule, but I don't really know how to do that. Please help me with this limit.
calculus limits
calculus limits
asked Jan 16 at 1:39
SimranSimran
153
asked Jan 16 at 1:39
SimranSimran
153
asked Jan 16 at 1:39
SimranSimran
153
asked Jan 16 at 1:39
SimranSimran
153
asked Jan 16 at 1:39
SimranSimran
153
153
$begingroup$
Note that L'Hopitals rule only applies to the $frac{0}{0}$ and $frac{infty}{infty}$ indeterminate forms.
$endgroup$
– H Huang
Jan 16 at 1:42
$begingroup$
Hint: $x = 1 + frac{1}{frac{1}{(x-1)}}$
$endgroup$
– user626177
Jan 16 at 1:43
add a comment |
$begingroup$
Note that L'Hopitals rule only applies to the $frac{0}{0}$ and $frac{infty}{infty}$ indeterminate forms.
$endgroup$
– H Huang
Jan 16 at 1:42
$begingroup$
Hint: $x = 1 + frac{1}{frac{1}{(x-1)}}$
$endgroup$
– user626177
Jan 16 at 1:43
$begingroup$
Note that L'Hopitals rule only applies to the $frac{0}{0}$ and $frac{infty}{infty}$ indeterminate forms.
$endgroup$
– H Huang
Jan 16 at 1:42
$begingroup$
Note that L'Hopitals rule only applies to the $frac{0}{0}$ and $frac{infty}{infty}$ indeterminate forms.
$endgroup$
– H Huang
Jan 16 at 1:42
$begingroup$
Hint: $x = 1 + frac{1}{frac{1}{(x-1)}}$
$endgroup$
– user626177
Jan 16 at 1:43
$begingroup$
Hint: $x = 1 + frac{1}{frac{1}{(x-1)}}$
$endgroup$
– user626177
Jan 16 at 1:43
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
$x^{{1over{x-1}}}$
$=e^{{{ln(x)}over{x-1}}}$
$lim_{xrightarrow 1}{{ln(x)}over{x-1}}$ is the derivative of $ln(x)$ at $1$ which is $1$. So the limit is $e$.
answered Jan 16 at 1:42
Tsemo AristideTsemo Aristide
58.4k11445
$endgroup$
add a comment |
$begingroup$
Set $x-1=h$ to find $$lim_{hto0}(1+h)^{1/h}=lim_{ntoinfty}left(1+dfrac1nright)^n=?$$
answered Jan 16 at 1:41
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
$begingroup$
Let $y = x^{(1/(x-1))}$
Take the natural log of both sides.
$ln(y) = (1/(x-1))ln(x)$
Notice if you plug in $x = 1$ you get $0/0$, so you can apply L'Hospital's Rule.
Just take the derivative of the numerator and denominator (separately). The derivative of $ln(x) = 1/x$ and the derivative of $x-1$ is just $1$.
So $ln(y) = 1/x$.
Now this is an easy limit, $ln(y) = 1$.
But that's not the limit you want.
Apply the exponential function to both sides.
Then you see the limit is $e$.
Hope this helps!
edited Jan 16 at 2:21
Larry
2,41331129
answered Jan 16 at 1:54
SydneySydney
234
$endgroup$
add a comment |
$begingroup$
Consider $$y=x^{frac{1}{x-1}}$$ and let $x=1+t$ to make
$$y=(1+t)^{frac{1}{t}}implies log(y)=frac 1 t log(1+t)$$ which is simple using L'Hospital.
If you want more than the limit, use Taylor series to get
$$log(y)=frac 1 t left(t-frac{t^2}{2}+frac{t^3}{3}+Oleft(t^4right) right)=1-frac{t}{2}+frac{t^2}{3}+Oleft(t^3right)$$ Continue with Taylor using
$$y=e^{log(y)}implies y=e-frac{e t}{2}+frac{11 e t^2}{24}+Oleft(t^3right)$$ which shows the limit when $t to 0$ and also how it is approached.
answered Jan 16 at 4:03
Claude LeiboviciClaude Leibovici
122k1157134
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$x^{{1over{x-1}}}$
$=e^{{{ln(x)}over{x-1}}}$
$lim_{xrightarrow 1}{{ln(x)}over{x-1}}$ is the derivative of $ln(x)$ at $1$ which is $1$. So the limit is $e$.
answered Jan 16 at 1:42
Tsemo AristideTsemo Aristide
58.4k11445
$endgroup$
add a comment |
$begingroup$
$x^{{1over{x-1}}}$
$=e^{{{ln(x)}over{x-1}}}$
$lim_{xrightarrow 1}{{ln(x)}over{x-1}}$ is the derivative of $ln(x)$ at $1$ which is $1$. So the limit is $e$.
answered Jan 16 at 1:42
Tsemo AristideTsemo Aristide
58.4k11445
$endgroup$
add a comment |
$begingroup$
$x^{{1over{x-1}}}$
$=e^{{{ln(x)}over{x-1}}}$
$lim_{xrightarrow 1}{{ln(x)}over{x-1}}$ is the derivative of $ln(x)$ at $1$ which is $1$. So the limit is $e$.
answered Jan 16 at 1:42
Tsemo AristideTsemo Aristide
58.4k11445
$endgroup$
$x^{{1over{x-1}}}$
$=e^{{{ln(x)}over{x-1}}}$
$lim_{xrightarrow 1}{{ln(x)}over{x-1}}$ is the derivative of $ln(x)$ at $1$ which is $1$. So the limit is $e$.
answered Jan 16 at 1:42
Tsemo AristideTsemo Aristide
58.4k11445
answered Jan 16 at 1:42
Tsemo AristideTsemo Aristide
58.4k11445
answered Jan 16 at 1:42
Tsemo AristideTsemo Aristide
58.4k11445
answered Jan 16 at 1:42
Tsemo AristideTsemo Aristide
58.4k11445
58.4k11445
add a comment |
add a comment |
$begingroup$
Set $x-1=h$ to find $$lim_{hto0}(1+h)^{1/h}=lim_{ntoinfty}left(1+dfrac1nright)^n=?$$
answered Jan 16 at 1:41
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
$begingroup$
Set $x-1=h$ to find $$lim_{hto0}(1+h)^{1/h}=lim_{ntoinfty}left(1+dfrac1nright)^n=?$$
answered Jan 16 at 1:41
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
$begingroup$
Set $x-1=h$ to find $$lim_{hto0}(1+h)^{1/h}=lim_{ntoinfty}left(1+dfrac1nright)^n=?$$
answered Jan 16 at 1:41
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
Set $x-1=h$ to find $$lim_{hto0}(1+h)^{1/h}=lim_{ntoinfty}left(1+dfrac1nright)^n=?$$
answered Jan 16 at 1:41
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 16 at 1:41
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 16 at 1:41
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 16 at 1:41
lab bhattacharjeelab bhattacharjee
226k15157275
226k15157275
add a comment |
add a comment |
$begingroup$
Let $y = x^{(1/(x-1))}$
Take the natural log of both sides.
$ln(y) = (1/(x-1))ln(x)$
Notice if you plug in $x = 1$ you get $0/0$, so you can apply L'Hospital's Rule.
Just take the derivative of the numerator and denominator (separately). The derivative of $ln(x) = 1/x$ and the derivative of $x-1$ is just $1$.
So $ln(y) = 1/x$.
Now this is an easy limit, $ln(y) = 1$.
But that's not the limit you want.
Apply the exponential function to both sides.
Then you see the limit is $e$.
Hope this helps!
edited Jan 16 at 2:21
Larry
2,41331129
answered Jan 16 at 1:54
SydneySydney
234
$endgroup$
add a comment |
$begingroup$
Let $y = x^{(1/(x-1))}$
Take the natural log of both sides.
$ln(y) = (1/(x-1))ln(x)$
Notice if you plug in $x = 1$ you get $0/0$, so you can apply L'Hospital's Rule.
Just take the derivative of the numerator and denominator (separately). The derivative of $ln(x) = 1/x$ and the derivative of $x-1$ is just $1$.
So $ln(y) = 1/x$.
Now this is an easy limit, $ln(y) = 1$.
But that's not the limit you want.
Apply the exponential function to both sides.
Then you see the limit is $e$.
Hope this helps!
edited Jan 16 at 2:21
Larry
2,41331129
answered Jan 16 at 1:54
SydneySydney
234
$endgroup$
add a comment |
$begingroup$
Let $y = x^{(1/(x-1))}$
Take the natural log of both sides.
$ln(y) = (1/(x-1))ln(x)$
Notice if you plug in $x = 1$ you get $0/0$, so you can apply L'Hospital's Rule.
Just take the derivative of the numerator and denominator (separately). The derivative of $ln(x) = 1/x$ and the derivative of $x-1$ is just $1$.
So $ln(y) = 1/x$.
Now this is an easy limit, $ln(y) = 1$.
But that's not the limit you want.
Apply the exponential function to both sides.
Then you see the limit is $e$.
Hope this helps!
edited Jan 16 at 2:21
Larry
2,41331129
answered Jan 16 at 1:54
SydneySydney
234
$endgroup$
Let $y = x^{(1/(x-1))}$
Take the natural log of both sides.
$ln(y) = (1/(x-1))ln(x)$
Notice if you plug in $x = 1$ you get $0/0$, so you can apply L'Hospital's Rule.
Just take the derivative of the numerator and denominator (separately). The derivative of $ln(x) = 1/x$ and the derivative of $x-1$ is just $1$.
So $ln(y) = 1/x$.
Now this is an easy limit, $ln(y) = 1$.
But that's not the limit you want.
Apply the exponential function to both sides.
Then you see the limit is $e$.
Hope this helps!
edited Jan 16 at 2:21
Larry
2,41331129
answered Jan 16 at 1:54
SydneySydney
234
edited Jan 16 at 2:21
Larry
2,41331129
edited Jan 16 at 2:21
Larry
2,41331129
edited Jan 16 at 2:21
Larry
2,41331129
2,41331129
answered Jan 16 at 1:54
SydneySydney
234
answered Jan 16 at 1:54
SydneySydney
234
answered Jan 16 at 1:54
SydneySydney
234
234
add a comment |
add a comment |
$begingroup$
Consider $$y=x^{frac{1}{x-1}}$$ and let $x=1+t$ to make
$$y=(1+t)^{frac{1}{t}}implies log(y)=frac 1 t log(1+t)$$ which is simple using L'Hospital.
If you want more than the limit, use Taylor series to get
$$log(y)=frac 1 t left(t-frac{t^2}{2}+frac{t^3}{3}+Oleft(t^4right) right)=1-frac{t}{2}+frac{t^2}{3}+Oleft(t^3right)$$ Continue with Taylor using
$$y=e^{log(y)}implies y=e-frac{e t}{2}+frac{11 e t^2}{24}+Oleft(t^3right)$$ which shows the limit when $t to 0$ and also how it is approached.
answered Jan 16 at 4:03
Claude LeiboviciClaude Leibovici
122k1157134
$endgroup$
add a comment |
$begingroup$
Consider $$y=x^{frac{1}{x-1}}$$ and let $x=1+t$ to make
$$y=(1+t)^{frac{1}{t}}implies log(y)=frac 1 t log(1+t)$$ which is simple using L'Hospital.
If you want more than the limit, use Taylor series to get
$$log(y)=frac 1 t left(t-frac{t^2}{2}+frac{t^3}{3}+Oleft(t^4right) right)=1-frac{t}{2}+frac{t^2}{3}+Oleft(t^3right)$$ Continue with Taylor using
$$y=e^{log(y)}implies y=e-frac{e t}{2}+frac{11 e t^2}{24}+Oleft(t^3right)$$ which shows the limit when $t to 0$ and also how it is approached.
answered Jan 16 at 4:03
Claude LeiboviciClaude Leibovici
122k1157134
$endgroup$
add a comment |
$begingroup$
Consider $$y=x^{frac{1}{x-1}}$$ and let $x=1+t$ to make
$$y=(1+t)^{frac{1}{t}}implies log(y)=frac 1 t log(1+t)$$ which is simple using L'Hospital.
If you want more than the limit, use Taylor series to get
$$log(y)=frac 1 t left(t-frac{t^2}{2}+frac{t^3}{3}+Oleft(t^4right) right)=1-frac{t}{2}+frac{t^2}{3}+Oleft(t^3right)$$ Continue with Taylor using
$$y=e^{log(y)}implies y=e-frac{e t}{2}+frac{11 e t^2}{24}+Oleft(t^3right)$$ which shows the limit when $t to 0$ and also how it is approached.
answered Jan 16 at 4:03
Claude LeiboviciClaude Leibovici
122k1157134
$endgroup$
Consider $$y=x^{frac{1}{x-1}}$$ and let $x=1+t$ to make
$$y=(1+t)^{frac{1}{t}}implies log(y)=frac 1 t log(1+t)$$ which is simple using L'Hospital.
If you want more than the limit, use Taylor series to get
$$log(y)=frac 1 t left(t-frac{t^2}{2}+frac{t^3}{3}+Oleft(t^4right) right)=1-frac{t}{2}+frac{t^2}{3}+Oleft(t^3right)$$ Continue with Taylor using
$$y=e^{log(y)}implies y=e-frac{e t}{2}+frac{11 e t^2}{24}+Oleft(t^3right)$$ which shows the limit when $t to 0$ and also how it is approached.
answered Jan 16 at 4:03
Claude LeiboviciClaude Leibovici
122k1157134
answered Jan 16 at 4:03
Claude LeiboviciClaude Leibovici
122k1157134
answered Jan 16 at 4:03
Claude LeiboviciClaude Leibovici
122k1157134
answered Jan 16 at 4:03
Claude LeiboviciClaude Leibovici
122k1157134
122k1157134
add a comment |
add a comment |
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$begingroup$
Note that L'Hopitals rule only applies to the $frac{0}{0}$ and $frac{infty}{infty}$ indeterminate forms.
$endgroup$
– H Huang
Jan 16 at 1:42
$begingroup$
Hint: $x = 1 + frac{1}{frac{1}{(x-1)}}$
$endgroup$
– user626177
Jan 16 at 1:43