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How do I solve $ lim_x to 0left[1 - xarctanleft(nxright)right]^, 1/x^2 $? [on hold]

3

1

$begingroup$

How do I solve:

$$
lim_x to 0left[1 - xarctanleft(nxright)right]^, 1/x^2
$$

I know the answer is $e^-n$ for every n > 1, but for the life of me I have no idea how to actually get to that answer. Am I supposed to use a common limit or any theorem? Also, why isn't the limit equal to 1?

Thanks.

calculus limits trigonometry exponentiation

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edited yesterday

user21820

39.8k544158

asked 2 days ago

radooradoo

184

New contributor

radoo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

put on hold as off-topic by user21820, RRL, José Carlos Santos, Adrian Keister, Cesareo 20 hours ago

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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, RRL, José Carlos Santos, Adrian Keister, Cesareo

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I accidentally put k instead of n, sorry.
  $endgroup$
  – radoo
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can take the logarithm of the function and use L'Hopitals rule
  $endgroup$
  – Nimish
  2 days ago
  

  
  
  
  

add a comment | 

3

1

$begingroup$

How do I solve:

$$
lim_x to 0left[1 - xarctanleft(nxright)right]^, 1/x^2
$$

I know the answer is $e^-n$ for every n > 1, but for the life of me I have no idea how to actually get to that answer. Am I supposed to use a common limit or any theorem? Also, why isn't the limit equal to 1?

Thanks.

calculus limits trigonometry exponentiation

share|cite|improve this question

edited yesterday

user21820

39.8k544158

asked 2 days ago

radooradoo

184

New contributor

radoo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

put on hold as off-topic by user21820, RRL, José Carlos Santos, Adrian Keister, Cesareo 20 hours ago

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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, RRL, José Carlos Santos, Adrian Keister, Cesareo

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I accidentally put k instead of n, sorry.
  $endgroup$
  – radoo
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can take the logarithm of the function and use L'Hopitals rule
  $endgroup$
  – Nimish
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

How do I solve:

$$
lim_x to 0left[1 - xarctanleft(nxright)right]^, 1/x^2
$$

I know the answer is $e^-n$ for every n > 1, but for the life of me I have no idea how to actually get to that answer. Am I supposed to use a common limit or any theorem? Also, why isn't the limit equal to 1?

Thanks.

calculus limits trigonometry exponentiation

share|cite|improve this question

edited yesterday

user21820

39.8k544158

asked 2 days ago

radooradoo

184

New contributor

radoo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited yesterday

user21820

39.8k544158

asked 2 days ago

radooradoo

184

New contributor

radoo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited yesterday

user21820

39.8k544158

asked 2 days ago

radooradoo

184

New contributor

radoo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited yesterday

user21820

39.8k544158

edited yesterday

user21820

39.8k544158

asked 2 days ago

radooradoo

184

New contributor

radoo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 2 days ago

radooradoo

184

3 Answers
3

active

oldest

votes

6

$begingroup$

You can do it the following way:

$$
lim_xto 0 (1-xarctan(nx))^1/x^2 = lim_xto 0 e^log((1-xarctan(nx))^1/x^2)
$$

doing some algebra on the exponent:
$$
lim_xto 0 exp(log((1-xarctan(nx))^1/x^2)) = lim_xto 0 expleft(fraclog((1-xarctan(nx))x^2)right)
$$

by limit rules:

$$
lim_xto 0 expleft(fraclog((1-xarctan(nx))x^2)right) =expleft[ lim_xto 0 left(fraclog((1-xarctan(nx))x^2)right)right]
$$

apply L'Hospital's rule and after some algebra you should get:
$$
expleft[- fracn+lim_xto 0 fracarctan(nx)x+n^2lim_xto 0 xarctan(nx)2right]
$$

simplifying

$$
expleft(-fracn+n+n^22right) = e^-n
$$

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answered yesterday

DashiDashi

758311

$endgroup$

add a comment | 

2

$begingroup$

Hint: $lim_xto 0 (1-nx)^(1/x)=e^-n$

Also look at the Taylor expansion of the arctan

share|cite|improve this answer

answered 2 days ago

A. PA. P

1386

$endgroup$

add a comment | 

0

$begingroup$

Noting that $u=xarctan nxto0$ and $(arctan nx)/xto n$ as $xto 0$, we have

$$(1-xarctan nx)^1/x^2=((1-xarctan nx)^1/(xarctan nx))^(arctan nx)/x=((1-u)^1/u)^(arctan nx)/xto (e^-1)^n=e^-n$$

using the general limit property $lim f(x)^g(x)=(lim f(x))^lim g(x)$, provided $lim f(x)$ and $lim g(x)$ both exist (with $lim f(x)ge0$) and are not both $0$.

share|cite|improve this answer

answered yesterday

Barry CipraBarry Cipra

60.5k655129

$endgroup$

add a comment | 

6

$begingroup$

You can do it the following way:

$$
lim_xto 0 (1-xarctan(nx))^1/x^2 = lim_xto 0 e^log((1-xarctan(nx))^1/x^2)
$$

doing some algebra on the exponent:
$$
lim_xto 0 exp(log((1-xarctan(nx))^1/x^2)) = lim_xto 0 expleft(fraclog((1-xarctan(nx))x^2)right)
$$

by limit rules:

$$
lim_xto 0 expleft(fraclog((1-xarctan(nx))x^2)right) =expleft[ lim_xto 0 left(fraclog((1-xarctan(nx))x^2)right)right]
$$

apply L'Hospital's rule and after some algebra you should get:
$$
expleft[- fracn+lim_xto 0 fracarctan(nx)x+n^2lim_xto 0 xarctan(nx)2right]
$$

simplifying

$$
expleft(-fracn+n+n^22right) = e^-n
$$

share|cite|improve this answer

answered yesterday

DashiDashi

758311

$endgroup$

add a comment | 

6

$begingroup$

You can do it the following way:

$$
lim_xto 0 (1-xarctan(nx))^1/x^2 = lim_xto 0 e^log((1-xarctan(nx))^1/x^2)
$$

doing some algebra on the exponent:
$$
lim_xto 0 exp(log((1-xarctan(nx))^1/x^2)) = lim_xto 0 expleft(fraclog((1-xarctan(nx))x^2)right)
$$

by limit rules:

$$
lim_xto 0 expleft(fraclog((1-xarctan(nx))x^2)right) =expleft[ lim_xto 0 left(fraclog((1-xarctan(nx))x^2)right)right]
$$

apply L'Hospital's rule and after some algebra you should get:
$$
expleft[- fracn+lim_xto 0 fracarctan(nx)x+n^2lim_xto 0 xarctan(nx)2right]
$$

simplifying

$$
expleft(-fracn+n+n^22right) = e^-n
$$

share|cite|improve this answer

answered yesterday

DashiDashi

758311

$endgroup$

add a comment | 

$begingroup$

You can do it the following way:

$$
lim_xto 0 (1-xarctan(nx))^1/x^2 = lim_xto 0 e^log((1-xarctan(nx))^1/x^2)
$$

doing some algebra on the exponent:
$$
lim_xto 0 exp(log((1-xarctan(nx))^1/x^2)) = lim_xto 0 expleft(fraclog((1-xarctan(nx))x^2)right)
$$

by limit rules:

$$
lim_xto 0 expleft(fraclog((1-xarctan(nx))x^2)right) =expleft[ lim_xto 0 left(fraclog((1-xarctan(nx))x^2)right)right]
$$

apply L'Hospital's rule and after some algebra you should get:
$$
expleft[- fracn+lim_xto 0 fracarctan(nx)x+n^2lim_xto 0 xarctan(nx)2right]
$$

simplifying

$$
expleft(-fracn+n+n^22right) = e^-n
$$

share|cite|improve this answer

answered yesterday

DashiDashi

758311

$endgroup$

share|cite|improve this answer

answered yesterday

DashiDashi

758311

answered yesterday

DashiDashi

758311

answered yesterday

DashiDashi

758311

2

$begingroup$

Hint: $lim_xto 0 (1-nx)^(1/x)=e^-n$

Also look at the Taylor expansion of the arctan

share|cite|improve this answer

answered 2 days ago

A. PA. P

1386

$endgroup$

add a comment | 

2

$begingroup$

Hint: $lim_xto 0 (1-nx)^(1/x)=e^-n$

Also look at the Taylor expansion of the arctan

share|cite|improve this answer

answered 2 days ago

A. PA. P

1386

$endgroup$

add a comment | 

$begingroup$

Hint: $lim_xto 0 (1-nx)^(1/x)=e^-n$

Also look at the Taylor expansion of the arctan

share|cite|improve this answer

answered 2 days ago

A. PA. P

1386

$endgroup$

share|cite|improve this answer

answered 2 days ago

A. PA. P

1386

answered 2 days ago

A. PA. P

1386

answered 2 days ago

A. PA. P

1386

0

$begingroup$

Noting that $u=xarctan nxto0$ and $(arctan nx)/xto n$ as $xto 0$, we have

$$(1-xarctan nx)^1/x^2=((1-xarctan nx)^1/(xarctan nx))^(arctan nx)/x=((1-u)^1/u)^(arctan nx)/xto (e^-1)^n=e^-n$$

using the general limit property $lim f(x)^g(x)=(lim f(x))^lim g(x)$, provided $lim f(x)$ and $lim g(x)$ both exist (with $lim f(x)ge0$) and are not both $0$.

share|cite|improve this answer

answered yesterday

Barry CipraBarry Cipra

60.5k655129

$endgroup$

add a comment | 

0

$begingroup$

Noting that $u=xarctan nxto0$ and $(arctan nx)/xto n$ as $xto 0$, we have

$$(1-xarctan nx)^1/x^2=((1-xarctan nx)^1/(xarctan nx))^(arctan nx)/x=((1-u)^1/u)^(arctan nx)/xto (e^-1)^n=e^-n$$

using the general limit property $lim f(x)^g(x)=(lim f(x))^lim g(x)$, provided $lim f(x)$ and $lim g(x)$ both exist (with $lim f(x)ge0$) and are not both $0$.

share|cite|improve this answer

answered yesterday

Barry CipraBarry Cipra

60.5k655129

$endgroup$

add a comment | 

$begingroup$

Noting that $u=xarctan nxto0$ and $(arctan nx)/xto n$ as $xto 0$, we have

$$(1-xarctan nx)^1/x^2=((1-xarctan nx)^1/(xarctan nx))^(arctan nx)/x=((1-u)^1/u)^(arctan nx)/xto (e^-1)^n=e^-n$$

using the general limit property $lim f(x)^g(x)=(lim f(x))^lim g(x)$, provided $lim f(x)$ and $lim g(x)$ both exist (with $lim f(x)ge0$) and are not both $0$.

share|cite|improve this answer

answered yesterday

Barry CipraBarry Cipra

60.5k655129

$endgroup$

share|cite|improve this answer

answered yesterday

Barry CipraBarry Cipra

60.5k655129

answered yesterday

Barry CipraBarry Cipra

60.5k655129

answered yesterday

Barry CipraBarry Cipra

60.5k655129