Evaluating $lim_xto0(1-x)^1/x$ [closed]

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Knowing
$$lim_xto0(1+x)^1/x=e$$
is it possible to evaluate or is it obvious that
$$lim_xto0(1-x)^1/x=e^-1$$

limits

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edited Jul 16 at 7:51

Parcly Taxel

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asked Jul 16 at 7:50

GRANZER

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closed as off-topic by Alex Francisco, John Ma, Strants, Mostafa Ayaz, Taroccoesbrocco Jul 16 at 19:04

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up vote
-1
down vote

favorite

Knowing
$$lim_xto0(1+x)^1/x=e$$
is it possible to evaluate or is it obvious that
$$lim_xto0(1-x)^1/x=e^-1$$

limits

share|cite|improve this question

edited Jul 16 at 7:51

Parcly Taxel

33.6k136588

asked Jul 16 at 7:50

GRANZER

1869

closed as off-topic by Alex Francisco, John Ma, Strants, Mostafa Ayaz, Taroccoesbrocco Jul 16 at 19:04

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, John Ma, Strants, Mostafa Ayaz, Taroccoesbrocco

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

up vote
-1
down vote

favorite

up vote
-1
down vote

favorite

Knowing
$$lim_xto0(1+x)^1/x=e$$
is it possible to evaluate or is it obvious that
$$lim_xto0(1-x)^1/x=e^-1$$

limits

share|cite|improve this question

edited Jul 16 at 7:51

Parcly Taxel

33.6k136588

asked Jul 16 at 7:50

GRANZER

1869

Knowing
$$lim_xto0(1+x)^1/x=e$$
is it possible to evaluate or is it obvious that
$$lim_xto0(1-x)^1/x=e^-1$$

limits

share|cite|improve this question

edited Jul 16 at 7:51

Parcly Taxel

33.6k136588

asked Jul 16 at 7:50

GRANZER

1869

share|cite|improve this question

share|cite|improve this question

edited Jul 16 at 7:51

Parcly Taxel

33.6k136588

edited Jul 16 at 7:51

Parcly Taxel

33.6k136588

edited Jul 16 at 7:51

Parcly Taxel

33.6k136588

33.6k136588

asked Jul 16 at 7:50

GRANZER

1869

asked Jul 16 at 7:50

GRANZER

1869

asked Jul 16 at 7:50

GRANZER

1869

1869

closed as off-topic by Alex Francisco, John Ma, Strants, Mostafa Ayaz, Taroccoesbrocco Jul 16 at 19:04

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, John Ma, Strants, Mostafa Ayaz, Taroccoesbrocco

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Alex Francisco, John Ma, Strants, Mostafa Ayaz, Taroccoesbrocco Jul 16 at 19:04

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, John Ma, Strants, Mostafa Ayaz, Taroccoesbrocco

If this question can be reworded to fit the rules in the help center, please edit the question.

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3 Answers
3

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oldest

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up vote
4
down vote



accepted

You can see the second limit as:

$displaystylelim_xto 0 frac1(1-x)^-1/x$

Then you use your first limit to get the asked for equality.

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answered Jul 16 at 7:55

Benoit Gaudeul

4208

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up vote
2
down vote

Guide:

• Do a substitution $y=-x$ and $frac1x$ is a continuous function.

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answered Jul 16 at 7:53

Siong Thye Goh

77.8k134796

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up vote
1
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Not quite. Use L'hopital's rule. We see it approaches the indeterminate form $1^infty$. So, taking the natural log, we see this limit is equal to $e^lim_xto 0log(1-x)^1/x$. We see $lim_xto 0log(1-x)^1/x=lim_xto 0fraclog(1-x)x$. Once again, this approaches the indeterminate form $frac00$ so we take the derivative of the top and bottom to get $lim_xto 0frac-11-x=-1$. Thus, the solution is $e^-1$. Alternatively, you could make a subsitution $y=-x$ and use the already known limit to obtain the same result.

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answered Jul 16 at 7:59

Julian Benali

27212

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote



accepted

You can see the second limit as:

$displaystylelim_xto 0 frac1(1-x)^-1/x$

Then you use your first limit to get the asked for equality.

share|cite|improve this answer

answered Jul 16 at 7:55

Benoit Gaudeul

4208

add a comment | 

up vote
4
down vote



accepted

You can see the second limit as:

$displaystylelim_xto 0 frac1(1-x)^-1/x$

Then you use your first limit to get the asked for equality.

share|cite|improve this answer

answered Jul 16 at 7:55

Benoit Gaudeul

4208

add a comment | 

up vote
4
down vote



accepted

up vote
4
down vote



accepted

You can see the second limit as:

$displaystylelim_xto 0 frac1(1-x)^-1/x$

Then you use your first limit to get the asked for equality.

share|cite|improve this answer

answered Jul 16 at 7:55

Benoit Gaudeul

4208

You can see the second limit as:

$displaystylelim_xto 0 frac1(1-x)^-1/x$

Then you use your first limit to get the asked for equality.

share|cite|improve this answer

answered Jul 16 at 7:55

Benoit Gaudeul

4208

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 7:55

Benoit Gaudeul

4208

answered Jul 16 at 7:55

Benoit Gaudeul

4208

answered Jul 16 at 7:55

Benoit Gaudeul

4208

4208

add a comment | 

add a comment | 

up vote
2
down vote

Guide:

• Do a substitution $y=-x$ and $frac1x$ is a continuous function.

share|cite|improve this answer

answered Jul 16 at 7:53

Siong Thye Goh

77.8k134796

add a comment | 

up vote
2
down vote

Guide:

• Do a substitution $y=-x$ and $frac1x$ is a continuous function.

share|cite|improve this answer

answered Jul 16 at 7:53

Siong Thye Goh

77.8k134796

add a comment | 

up vote
2
down vote

up vote
2
down vote

Guide:

• Do a substitution $y=-x$ and $frac1x$ is a continuous function.

share|cite|improve this answer

answered Jul 16 at 7:53

Siong Thye Goh

77.8k134796

Guide:

• Do a substitution $y=-x$ and $frac1x$ is a continuous function.

share|cite|improve this answer

answered Jul 16 at 7:53

Siong Thye Goh

77.8k134796

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 7:53

Siong Thye Goh

77.8k134796

answered Jul 16 at 7:53

Siong Thye Goh

77.8k134796

answered Jul 16 at 7:53

Siong Thye Goh

77.8k134796

77.8k134796

add a comment | 

add a comment | 

up vote
1
down vote

Not quite. Use L'hopital's rule. We see it approaches the indeterminate form $1^infty$. So, taking the natural log, we see this limit is equal to $e^lim_xto 0log(1-x)^1/x$. We see $lim_xto 0log(1-x)^1/x=lim_xto 0fraclog(1-x)x$. Once again, this approaches the indeterminate form $frac00$ so we take the derivative of the top and bottom to get $lim_xto 0frac-11-x=-1$. Thus, the solution is $e^-1$. Alternatively, you could make a subsitution $y=-x$ and use the already known limit to obtain the same result.

share|cite|improve this answer

answered Jul 16 at 7:59

Julian Benali

27212

add a comment | 

up vote
1
down vote

Not quite. Use L'hopital's rule. We see it approaches the indeterminate form $1^infty$. So, taking the natural log, we see this limit is equal to $e^lim_xto 0log(1-x)^1/x$. We see $lim_xto 0log(1-x)^1/x=lim_xto 0fraclog(1-x)x$. Once again, this approaches the indeterminate form $frac00$ so we take the derivative of the top and bottom to get $lim_xto 0frac-11-x=-1$. Thus, the solution is $e^-1$. Alternatively, you could make a subsitution $y=-x$ and use the already known limit to obtain the same result.

share|cite|improve this answer

answered Jul 16 at 7:59

Julian Benali

27212

add a comment | 

up vote
1
down vote

up vote
1
down vote

Not quite. Use L'hopital's rule. We see it approaches the indeterminate form $1^infty$. So, taking the natural log, we see this limit is equal to $e^lim_xto 0log(1-x)^1/x$. We see $lim_xto 0log(1-x)^1/x=lim_xto 0fraclog(1-x)x$. Once again, this approaches the indeterminate form $frac00$ so we take the derivative of the top and bottom to get $lim_xto 0frac-11-x=-1$. Thus, the solution is $e^-1$. Alternatively, you could make a subsitution $y=-x$ and use the already known limit to obtain the same result.

share|cite|improve this answer

answered Jul 16 at 7:59

Julian Benali

27212

Not quite. Use L'hopital's rule. We see it approaches the indeterminate form $1^infty$. So, taking the natural log, we see this limit is equal to $e^lim_xto 0log(1-x)^1/x$. We see $lim_xto 0log(1-x)^1/x=lim_xto 0fraclog(1-x)x$. Once again, this approaches the indeterminate form $frac00$ so we take the derivative of the top and bottom to get $lim_xto 0frac-11-x=-1$. Thus, the solution is $e^-1$. Alternatively, you could make a subsitution $y=-x$ and use the already known limit to obtain the same result.

share|cite|improve this answer

answered Jul 16 at 7:59

Julian Benali

27212

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 7:59

Julian Benali

27212

answered Jul 16 at 7:59

Julian Benali

27212

answered Jul 16 at 7:59

Julian Benali

27212

27212

add a comment | 

add a comment |