How do I solve this limit? The Next CEO of Stack Overflowrational limit problemHow to evaluate the following limit? $limlimits_xtoinftyxleft(fracpi2-arctan xright).$Limit only using squeeze theorem $ lim_(x,y)to(0,2) x,arctanleft(frac1y-2right)$How to calculate this limit at infinity?How can I use the Limit Laws to solve this limit?How do I even start approaching this limit?how can I find this limitHow to solve this limit without L'Hospital?How to solve this limit when direct substitution fails. Why do this work?L'Hôpital's rule - How solve this limit question
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How do I solve this limit?
The Next CEO of Stack Overflowrational limit problemHow to evaluate the following limit? $limlimits_xtoinftyxleft(fracpi2-arctan xright).$Limit only using squeeze theorem $ lim_(x,y)to(0,2) x,arctanleft(frac1y-2right)$How to calculate this limit at infinity?How can I use the Limit Laws to solve this limit?How do I even start approaching this limit?how can I find this limitHow to solve this limit without L'Hospital?How to solve this limit when direct substitution fails. Why do this work?L'Hôpital's rule - How solve this limit question
$begingroup$
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
edited 47 mins ago
Felix Marin
68.8k7109146
asked 1 hour ago
radooradoo
84
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.
$endgroup$
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
edited 47 mins ago
Felix Marin
68.8k7109146
asked 1 hour ago
radooradoo
84
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.
$endgroup$
$begingroup$
I accidentally put k instead of n, sorry.
$endgroup$
– radoo
1 hour ago
$begingroup$
You can take the logarithm of the function and use L'Hopitals rule
$endgroup$
– Nimish
1 hour 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
edited 47 mins ago
Felix Marin
68.8k7109146
asked 1 hour ago
radooradoo
84
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.
$endgroup$
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
calculus limits
edited 47 mins ago
Felix Marin
68.8k7109146
asked 1 hour ago
radooradoo
84
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 47 mins ago
Felix Marin
68.8k7109146
asked 1 hour ago
radooradoo
84
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 47 mins ago
Felix Marin
68.8k7109146
edited 47 mins ago
Felix Marin
68.8k7109146
edited 47 mins ago
Felix Marin
68.8k7109146
68.8k7109146
asked 1 hour ago
radooradoo
84
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 1 hour ago
radooradoo
84
asked 1 hour ago
radooradoo
84
84
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.
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.
radoo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
I accidentally put k instead of n, sorry.
$endgroup$
– radoo
1 hour ago
$begingroup$
You can take the logarithm of the function and use L'Hopitals rule
$endgroup$
– Nimish
1 hour ago
add a comment |
$begingroup$
I accidentally put k instead of n, sorry.
$endgroup$
– radoo
1 hour ago
$begingroup$
You can take the logarithm of the function and use L'Hopitals rule
$endgroup$
– Nimish
1 hour ago
$begingroup$
I accidentally put k instead of n, sorry.
$endgroup$
– radoo
1 hour ago
$begingroup$
I accidentally put k instead of n, sorry.
$endgroup$
– radoo
1 hour ago
$begingroup$
You can take the logarithm of the function and use L'Hopitals rule
$endgroup$
– Nimish
1 hour ago
$begingroup$
You can take the logarithm of the function and use L'Hopitals rule
$endgroup$
– Nimish
1 hour ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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
$$
expleft(-fracn+n+n^22right) = e^-n
$$
answered 1 hour ago
DashiDashi
726311
$endgroup$
add a comment |
$begingroup$
Hint: $lim_xto 0 (1-nx)^(1/x)=e^-n$
Also look at the Taylor expansion of the arctan
answered 1 hour ago
A. PA. P
1386
$endgroup$
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 $0$.
answered 20 mins ago
Barry CipraBarry Cipra
60.5k655128
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
$$
expleft(-fracn+n+n^22right) = e^-n
$$
answered 1 hour ago
DashiDashi
726311
$endgroup$
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
$$
expleft(-fracn+n+n^22right) = e^-n
$$
answered 1 hour ago
DashiDashi
726311
$endgroup$
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
$$
expleft(-fracn+n+n^22right) = e^-n
$$
answered 1 hour ago
DashiDashi
726311
$endgroup$
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
$$
expleft(-fracn+n+n^22right) = e^-n
$$
answered 1 hour ago
DashiDashi
726311
answered 1 hour ago
DashiDashi
726311
answered 1 hour ago
DashiDashi
726311
answered 1 hour ago
DashiDashi
726311
726311
add a comment |
add a comment |
$begingroup$
Hint: $lim_xto 0 (1-nx)^(1/x)=e^-n$
Also look at the Taylor expansion of the arctan
answered 1 hour 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
answered 1 hour 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
answered 1 hour ago
A. PA. P
1386
$endgroup$
Hint: $lim_xto 0 (1-nx)^(1/x)=e^-n$
Also look at the Taylor expansion of the arctan
answered 1 hour ago
A. PA. P
1386
answered 1 hour ago
A. PA. P
1386
answered 1 hour ago
A. PA. P
1386
answered 1 hour ago
A. PA. P
1386
1386
add a comment |
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 $0$.
answered 20 mins ago
Barry CipraBarry Cipra
60.5k655128
$endgroup$
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 $0$.
answered 20 mins ago
Barry CipraBarry Cipra
60.5k655128
$endgroup$
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 $0$.
answered 20 mins ago
Barry CipraBarry Cipra
60.5k655128
$endgroup$
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 $0$.
answered 20 mins ago
Barry CipraBarry Cipra
60.5k655128
answered 20 mins ago
Barry CipraBarry Cipra
60.5k655128
answered 20 mins ago
Barry CipraBarry Cipra
60.5k655128
answered 20 mins ago
Barry CipraBarry Cipra
60.5k655128
60.5k655128
add a comment |
add a comment |
radoo is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
I accidentally put k instead of n, sorry.
$endgroup$
– radoo
1 hour ago
$begingroup$
You can take the logarithm of the function and use L'Hopitals rule
$endgroup$
– Nimish
1 hour ago