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?
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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 7 hours ago
Felix Marin
68.8k7109146
asked 8 hours 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.
$endgroup$
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 7 hours ago
Felix Marin
68.8k7109146
asked 8 hours 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.
$endgroup$
$begingroup$
I accidentally put k instead of n, sorry.
$endgroup$
– radoo
7 hours ago
$begingroup$
You can take the logarithm of the function and use L'Hopitals rule
$endgroup$
– Nimish
7 hours 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 7 hours ago
Felix Marin
68.8k7109146
asked 8 hours 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.
$endgroup$
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 7 hours ago
Felix Marin
68.8k7109146
asked 8 hours 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 7 hours ago
Felix Marin
68.8k7109146
asked 8 hours 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 7 hours ago
Felix Marin
68.8k7109146
edited 7 hours ago
Felix Marin
68.8k7109146
edited 7 hours ago
Felix Marin
68.8k7109146
68.8k7109146
asked 8 hours 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 8 hours ago
radooradoo
184
asked 8 hours ago
radooradoo
184
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.
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
7 hours ago
$begingroup$
You can take the logarithm of the function and use L'Hopitals rule
$endgroup$
– Nimish
7 hours ago
add a comment |
$begingroup$
I accidentally put k instead of n, sorry.
$endgroup$
– radoo
7 hours ago
$begingroup$
You can take the logarithm of the function and use L'Hopitals rule
$endgroup$
– Nimish
7 hours ago
$begingroup$
I accidentally put k instead of n, sorry.
$endgroup$
– radoo
7 hours ago
$begingroup$
I accidentally put k instead of n, sorry.
$endgroup$
– radoo
7 hours ago
$begingroup$
You can take the logarithm of the function and use L'Hopitals rule
$endgroup$
– Nimish
7 hours ago
$begingroup$
You can take the logarithm of the function and use L'Hopitals rule
$endgroup$
– Nimish
7 hours 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 7 hours ago
DashiDashi
746311
$endgroup$
add a comment |
$begingroup$
Hint: $lim_xto 0 (1-nx)^(1/x)=e^-n$
Also look at the Taylor expansion of the arctan
answered 7 hours 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 6 hours ago
Barry CipraBarry Cipra
60.5k655128
$endgroup$
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
$begingroup$
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 7 hours ago
DashiDashi
746311
$endgroup$
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 7 hours ago
DashiDashi
746311
$endgroup$
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 7 hours ago
DashiDashi
746311
$endgroup$
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 7 hours ago
DashiDashi
746311
answered 7 hours ago
DashiDashi
746311
answered 7 hours ago
DashiDashi
746311
answered 7 hours ago
DashiDashi
746311
746311
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 7 hours 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 7 hours 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 7 hours 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 7 hours ago
A. PA. P
1386
answered 7 hours ago
A. PA. P
1386
answered 7 hours ago
A. PA. P
1386
answered 7 hours 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 6 hours 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 6 hours 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 6 hours 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 6 hours ago
Barry CipraBarry Cipra
60.5k655128
answered 6 hours ago
Barry CipraBarry Cipra
60.5k655128
answered 6 hours ago
Barry CipraBarry Cipra
60.5k655128
answered 6 hours 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.
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
7 hours ago
$begingroup$
You can take the logarithm of the function and use L'Hopitals rule
$endgroup$
– Nimish
7 hours ago