Mixing
Computing limit of $f(x) = x^{(frac{1}{x} - 1)}$
Let $f(x) = x^{(frac{1}{x} - 1)}$ . Find $lim_{x to 0^{+}} f(x)$ if it exists .
My try : $f(x) = x^{(frac{1}{x} - 1)} = e^{frac{(1-x) ln x}{x}}$ . I'm not allowed to use L'Hôpital's rule but intuitively seems the answer is zero .
calculus real-analysis limits limits-without-lhopital
asked Nov 20 '18 at 0:42
S.H.W
1,1791922
add a comment |
Let $f(x) = x^{(frac{1}{x} - 1)}$ . Find $lim_{x to 0^{+}} f(x)$ if it exists .
My try : $f(x) = x^{(frac{1}{x} - 1)} = e^{frac{(1-x) ln x}{x}}$ . I'm not allowed to use L'Hôpital's rule but intuitively seems the answer is zero .
calculus real-analysis limits limits-without-lhopital
asked Nov 20 '18 at 0:42
S.H.W
1,1791922
The base tends to $0$ and exponent tends to $infty$ so the desired limit is $0$.
– Paramanand Singh
Nov 20 '18 at 2:01
@ParamanandSingh How we can prove that statement ?
– S.H.W
Nov 20 '18 at 3:16
Proving this is at the same level as proving that $1/xtoinfty$ as $xto 0^{+}$. Such proofs should not be demanded as a part of exercise but should rather form a part of your understanding. For that purpose let $0<x<1/2$ then $0<x^{(1/x)-1}<x$ and you can either use Squeeze Theorem or the $epsilon, delta $ definition of limit.
– Paramanand Singh
Nov 20 '18 at 5:03
@ParamanandSingh Thanks a lot .
– S.H.W
Nov 20 '18 at 17:49
add a comment |
Let $f(x) = x^{(frac{1}{x} - 1)}$ . Find $lim_{x to 0^{+}} f(x)$ if it exists .
My try : $f(x) = x^{(frac{1}{x} - 1)} = e^{frac{(1-x) ln x}{x}}$ . I'm not allowed to use L'Hôpital's rule but intuitively seems the answer is zero .
calculus real-analysis limits limits-without-lhopital
asked Nov 20 '18 at 0:42
S.H.W
1,1791922
Let $f(x) = x^{(frac{1}{x} - 1)}$ . Find $lim_{x to 0^{+}} f(x)$ if it exists .
My try : $f(x) = x^{(frac{1}{x} - 1)} = e^{frac{(1-x) ln x}{x}}$ . I'm not allowed to use L'Hôpital's rule but intuitively seems the answer is zero .
calculus real-analysis limits limits-without-lhopital
calculus real-analysis limits limits-without-lhopital
asked Nov 20 '18 at 0:42
S.H.W
1,1791922
asked Nov 20 '18 at 0:42
S.H.W
1,1791922
asked Nov 20 '18 at 0:42
S.H.W
1,1791922
asked Nov 20 '18 at 0:42
S.H.W
1,1791922
asked Nov 20 '18 at 0:42
S.H.W
1,1791922
1,1791922
The base tends to $0$ and exponent tends to $infty$ so the desired limit is $0$.
– Paramanand Singh
Nov 20 '18 at 2:01
@ParamanandSingh How we can prove that statement ?
– S.H.W
Nov 20 '18 at 3:16
Proving this is at the same level as proving that $1/xtoinfty$ as $xto 0^{+}$. Such proofs should not be demanded as a part of exercise but should rather form a part of your understanding. For that purpose let $0<x<1/2$ then $0<x^{(1/x)-1}<x$ and you can either use Squeeze Theorem or the $epsilon, delta $ definition of limit.
– Paramanand Singh
Nov 20 '18 at 5:03
@ParamanandSingh Thanks a lot .
– S.H.W
Nov 20 '18 at 17:49
add a comment |
The base tends to $0$ and exponent tends to $infty$ so the desired limit is $0$.
– Paramanand Singh
Nov 20 '18 at 2:01
@ParamanandSingh How we can prove that statement ?
– S.H.W
Nov 20 '18 at 3:16
Proving this is at the same level as proving that $1/xtoinfty$ as $xto 0^{+}$. Such proofs should not be demanded as a part of exercise but should rather form a part of your understanding. For that purpose let $0<x<1/2$ then $0<x^{(1/x)-1}<x$ and you can either use Squeeze Theorem or the $epsilon, delta $ definition of limit.
– Paramanand Singh
Nov 20 '18 at 5:03
@ParamanandSingh Thanks a lot .
– S.H.W
Nov 20 '18 at 17:49
The base tends to $0$ and exponent tends to $infty$ so the desired limit is $0$.
– Paramanand Singh
Nov 20 '18 at 2:01
The base tends to $0$ and exponent tends to $infty$ so the desired limit is $0$.
– Paramanand Singh
Nov 20 '18 at 2:01
@ParamanandSingh How we can prove that statement ?
– S.H.W
Nov 20 '18 at 3:16
@ParamanandSingh How we can prove that statement ?
– S.H.W
Nov 20 '18 at 3:16
Proving this is at the same level as proving that $1/xtoinfty$ as $xto 0^{+}$. Such proofs should not be demanded as a part of exercise but should rather form a part of your understanding. For that purpose let $0<x<1/2$ then $0<x^{(1/x)-1}<x$ and you can either use Squeeze Theorem or the $epsilon, delta $ definition of limit.
– Paramanand Singh
Nov 20 '18 at 5:03
Proving this is at the same level as proving that $1/xtoinfty$ as $xto 0^{+}$. Such proofs should not be demanded as a part of exercise but should rather form a part of your understanding. For that purpose let $0<x<1/2$ then $0<x^{(1/x)-1}<x$ and you can either use Squeeze Theorem or the $epsilon, delta $ definition of limit.
– Paramanand Singh
Nov 20 '18 at 5:03
@ParamanandSingh Thanks a lot .
– S.H.W
Nov 20 '18 at 17:49
@ParamanandSingh Thanks a lot .
– S.H.W
Nov 20 '18 at 17:49
add a comment |
3 Answers
3
active
oldest
votes
This is not indeterminate when $xto 0^+$:
$ln xto -infty$,
$1-xto 1$, so $(1-x)ln xto-infty$
- the denominator tends to $0^+$
so the exponent tends to $-infty$, and the exponential indeed tends to $0$.
answered Nov 20 '18 at 0:51
Bernard
118k639112
How we can make it in a rigorous way ?
– S.H.W
Nov 20 '18 at 0:52
add a comment |
$$frac{(1-x)log{x}}{x}=frac{1-x}{x}log{x}=(frac{1}{x}-1)log{x}$$
$$lim_{xto 0^{+}}{(frac{1}{x}-1)}=+infty text{ and } lim_{xto 0^{+}}{log{x}}=-infty$$
$$lim_{xto 0^{+}}{(frac{1}{x}-1)log{x}}=-infty$$
$$lim_{xto 0^{+}}{x^{frac{1}{x}-1}}=lim_{xto 0^{+}}{e^{frac{(1-x)log{x}}{x}}}=0$$
answered Nov 20 '18 at 1:54
Offlaw
2649
add a comment |
Hint: remember $infty times - infty= -infty.$Take $e$ and $ln$. Also, since $e$ is a continuous function so it behaves nicely with limit. I hope it should be clear now.
answered Nov 20 '18 at 2:09
gb2017
944
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
This is not indeterminate when $xto 0^+$:
$ln xto -infty$,
$1-xto 1$, so $(1-x)ln xto-infty$
- the denominator tends to $0^+$
so the exponent tends to $-infty$, and the exponential indeed tends to $0$.
answered Nov 20 '18 at 0:51
Bernard
118k639112
How we can make it in a rigorous way ?
– S.H.W
Nov 20 '18 at 0:52
add a comment |
This is not indeterminate when $xto 0^+$:
$ln xto -infty$,
$1-xto 1$, so $(1-x)ln xto-infty$
- the denominator tends to $0^+$
so the exponent tends to $-infty$, and the exponential indeed tends to $0$.
answered Nov 20 '18 at 0:51
Bernard
118k639112
How we can make it in a rigorous way ?
– S.H.W
Nov 20 '18 at 0:52
add a comment |
This is not indeterminate when $xto 0^+$:
$ln xto -infty$,
$1-xto 1$, so $(1-x)ln xto-infty$
- the denominator tends to $0^+$
so the exponent tends to $-infty$, and the exponential indeed tends to $0$.
answered Nov 20 '18 at 0:51
Bernard
118k639112
This is not indeterminate when $xto 0^+$:
$ln xto -infty$,
$1-xto 1$, so $(1-x)ln xto-infty$
- the denominator tends to $0^+$
so the exponent tends to $-infty$, and the exponential indeed tends to $0$.
answered Nov 20 '18 at 0:51
Bernard
118k639112
answered Nov 20 '18 at 0:51
Bernard
118k639112
answered Nov 20 '18 at 0:51
Bernard
118k639112
answered Nov 20 '18 at 0:51
Bernard
118k639112
118k639112
How we can make it in a rigorous way ?
– S.H.W
Nov 20 '18 at 0:52
add a comment |
How we can make it in a rigorous way ?
– S.H.W
Nov 20 '18 at 0:52
How we can make it in a rigorous way ?
– S.H.W
Nov 20 '18 at 0:52
How we can make it in a rigorous way ?
– S.H.W
Nov 20 '18 at 0:52
add a comment |
$$frac{(1-x)log{x}}{x}=frac{1-x}{x}log{x}=(frac{1}{x}-1)log{x}$$
$$lim_{xto 0^{+}}{(frac{1}{x}-1)}=+infty text{ and } lim_{xto 0^{+}}{log{x}}=-infty$$
$$lim_{xto 0^{+}}{(frac{1}{x}-1)log{x}}=-infty$$
$$lim_{xto 0^{+}}{x^{frac{1}{x}-1}}=lim_{xto 0^{+}}{e^{frac{(1-x)log{x}}{x}}}=0$$
answered Nov 20 '18 at 1:54
Offlaw
2649
add a comment |
$$frac{(1-x)log{x}}{x}=frac{1-x}{x}log{x}=(frac{1}{x}-1)log{x}$$
$$lim_{xto 0^{+}}{(frac{1}{x}-1)}=+infty text{ and } lim_{xto 0^{+}}{log{x}}=-infty$$
$$lim_{xto 0^{+}}{(frac{1}{x}-1)log{x}}=-infty$$
$$lim_{xto 0^{+}}{x^{frac{1}{x}-1}}=lim_{xto 0^{+}}{e^{frac{(1-x)log{x}}{x}}}=0$$
answered Nov 20 '18 at 1:54
Offlaw
2649
add a comment |
$$frac{(1-x)log{x}}{x}=frac{1-x}{x}log{x}=(frac{1}{x}-1)log{x}$$
$$lim_{xto 0^{+}}{(frac{1}{x}-1)}=+infty text{ and } lim_{xto 0^{+}}{log{x}}=-infty$$
$$lim_{xto 0^{+}}{(frac{1}{x}-1)log{x}}=-infty$$
$$lim_{xto 0^{+}}{x^{frac{1}{x}-1}}=lim_{xto 0^{+}}{e^{frac{(1-x)log{x}}{x}}}=0$$
answered Nov 20 '18 at 1:54
Offlaw
2649
$$frac{(1-x)log{x}}{x}=frac{1-x}{x}log{x}=(frac{1}{x}-1)log{x}$$
$$lim_{xto 0^{+}}{(frac{1}{x}-1)}=+infty text{ and } lim_{xto 0^{+}}{log{x}}=-infty$$
$$lim_{xto 0^{+}}{(frac{1}{x}-1)log{x}}=-infty$$
$$lim_{xto 0^{+}}{x^{frac{1}{x}-1}}=lim_{xto 0^{+}}{e^{frac{(1-x)log{x}}{x}}}=0$$
answered Nov 20 '18 at 1:54
Offlaw
2649
answered Nov 20 '18 at 1:54
Offlaw
2649
answered Nov 20 '18 at 1:54
Offlaw
2649
answered Nov 20 '18 at 1:54
Offlaw
2649
2649
add a comment |
add a comment |
Hint: remember $infty times - infty= -infty.$Take $e$ and $ln$. Also, since $e$ is a continuous function so it behaves nicely with limit. I hope it should be clear now.
answered Nov 20 '18 at 2:09
gb2017
944
add a comment |
Hint: remember $infty times - infty= -infty.$Take $e$ and $ln$. Also, since $e$ is a continuous function so it behaves nicely with limit. I hope it should be clear now.
answered Nov 20 '18 at 2:09
gb2017
944
add a comment |
Hint: remember $infty times - infty= -infty.$Take $e$ and $ln$. Also, since $e$ is a continuous function so it behaves nicely with limit. I hope it should be clear now.
answered Nov 20 '18 at 2:09
gb2017
944
Hint: remember $infty times - infty= -infty.$Take $e$ and $ln$. Also, since $e$ is a continuous function so it behaves nicely with limit. I hope it should be clear now.
answered Nov 20 '18 at 2:09
gb2017
944
answered Nov 20 '18 at 2:09
gb2017
944
answered Nov 20 '18 at 2:09
gb2017
944
answered Nov 20 '18 at 2:09
gb2017
944
944
add a comment |
add a comment |
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The base tends to $0$ and exponent tends to $infty$ so the desired limit is $0$.
– Paramanand Singh
Nov 20 '18 at 2:01
@ParamanandSingh How we can prove that statement ?
– S.H.W
Nov 20 '18 at 3:16
Proving this is at the same level as proving that $1/xtoinfty$ as $xto 0^{+}$. Such proofs should not be demanded as a part of exercise but should rather form a part of your understanding. For that purpose let $0<x<1/2$ then $0<x^{(1/x)-1}<x$ and you can either use Squeeze Theorem or the $epsilon, delta $ definition of limit.
– Paramanand Singh
Nov 20 '18 at 5:03
@ParamanandSingh Thanks a lot .
– S.H.W
Nov 20 '18 at 17:49