How to find $lim_{x to infty} (1+sinh(x^{-2}))^{(x^2)}$ [on hold]
I'm stuck with this task quite long, and I could use some new idea . The soultion should be e.
calculus limits limits-without-lhopital
New contributor
put on hold as unclear what you're asking by Mark Viola, amWhy, Mark Bennet, Davide Giraudo, Adrian Keister Jan 3 at 23:03
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
I'm stuck with this task quite long, and I could use some new idea . The soultion should be e.
calculus limits limits-without-lhopital
New contributor
put on hold as unclear what you're asking by Mark Viola, amWhy, Mark Bennet, Davide Giraudo, Adrian Keister Jan 3 at 23:03
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
What on earth is "sh?" Do you mean $sinh$, the hyperbolic sine function?
– Mark Viola
Jan 3 at 21:38
Yes, sorry, we use sh for sinh
– MilO
Jan 3 at 21:42
Hi! Welcome to MSE. Please use Mathjax wherever possible and provide a little more context as to why this problem is relevant to you and this community
– Sauhard Sharma
Jan 3 at 21:43
You don't want answers that use L'Hopital's rule?
– 0x539
Jan 3 at 21:53
add a comment |
I'm stuck with this task quite long, and I could use some new idea . The soultion should be e.
calculus limits limits-without-lhopital
New contributor
I'm stuck with this task quite long, and I could use some new idea . The soultion should be e.
calculus limits limits-without-lhopital
calculus limits limits-without-lhopital
New contributor
New contributor
edited Jan 3 at 21:50
0x539
1,047316
1,047316
New contributor
asked Jan 3 at 21:31
MilO
61
61
New contributor
New contributor
put on hold as unclear what you're asking by Mark Viola, amWhy, Mark Bennet, Davide Giraudo, Adrian Keister Jan 3 at 23:03
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as unclear what you're asking by Mark Viola, amWhy, Mark Bennet, Davide Giraudo, Adrian Keister Jan 3 at 23:03
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
What on earth is "sh?" Do you mean $sinh$, the hyperbolic sine function?
– Mark Viola
Jan 3 at 21:38
Yes, sorry, we use sh for sinh
– MilO
Jan 3 at 21:42
Hi! Welcome to MSE. Please use Mathjax wherever possible and provide a little more context as to why this problem is relevant to you and this community
– Sauhard Sharma
Jan 3 at 21:43
You don't want answers that use L'Hopital's rule?
– 0x539
Jan 3 at 21:53
add a comment |
1
What on earth is "sh?" Do you mean $sinh$, the hyperbolic sine function?
– Mark Viola
Jan 3 at 21:38
Yes, sorry, we use sh for sinh
– MilO
Jan 3 at 21:42
Hi! Welcome to MSE. Please use Mathjax wherever possible and provide a little more context as to why this problem is relevant to you and this community
– Sauhard Sharma
Jan 3 at 21:43
You don't want answers that use L'Hopital's rule?
– 0x539
Jan 3 at 21:53
1
1
What on earth is "sh?" Do you mean $sinh$, the hyperbolic sine function?
– Mark Viola
Jan 3 at 21:38
What on earth is "sh?" Do you mean $sinh$, the hyperbolic sine function?
– Mark Viola
Jan 3 at 21:38
Yes, sorry, we use sh for sinh
– MilO
Jan 3 at 21:42
Yes, sorry, we use sh for sinh
– MilO
Jan 3 at 21:42
Hi! Welcome to MSE. Please use Mathjax wherever possible and provide a little more context as to why this problem is relevant to you and this community
– Sauhard Sharma
Jan 3 at 21:43
Hi! Welcome to MSE. Please use Mathjax wherever possible and provide a little more context as to why this problem is relevant to you and this community
– Sauhard Sharma
Jan 3 at 21:43
You don't want answers that use L'Hopital's rule?
– 0x539
Jan 3 at 21:53
You don't want answers that use L'Hopital's rule?
– 0x539
Jan 3 at 21:53
add a comment |
2 Answers
2
active
oldest
votes
The usual idea is to compute
$$
l=lim_{xtoinfty}logbigl((1+sinh(x^{-2}))^{x^2}bigr)
$$
and then your limit will be $e^l$. Now, with the substitution $t=1/x^2$, you get
$$
lim_{tto0^+}frac{log(1+sinh t)}{t}
$$
add a comment |
This limit becomes obvious if you recall that $sinh x sim x$ as $x to 0$ (just as regular $sin$ is). So substitute $t = {1over x}$:
$$
lim_{tto 0}(1+sinh(t^2))^{1over t^2} sim lim_{tto 0}(1+t^2)^{1over t^2}
$$
RHS is a very famous limit.
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The usual idea is to compute
$$
l=lim_{xtoinfty}logbigl((1+sinh(x^{-2}))^{x^2}bigr)
$$
and then your limit will be $e^l$. Now, with the substitution $t=1/x^2$, you get
$$
lim_{tto0^+}frac{log(1+sinh t)}{t}
$$
add a comment |
The usual idea is to compute
$$
l=lim_{xtoinfty}logbigl((1+sinh(x^{-2}))^{x^2}bigr)
$$
and then your limit will be $e^l$. Now, with the substitution $t=1/x^2$, you get
$$
lim_{tto0^+}frac{log(1+sinh t)}{t}
$$
add a comment |
The usual idea is to compute
$$
l=lim_{xtoinfty}logbigl((1+sinh(x^{-2}))^{x^2}bigr)
$$
and then your limit will be $e^l$. Now, with the substitution $t=1/x^2$, you get
$$
lim_{tto0^+}frac{log(1+sinh t)}{t}
$$
The usual idea is to compute
$$
l=lim_{xtoinfty}logbigl((1+sinh(x^{-2}))^{x^2}bigr)
$$
and then your limit will be $e^l$. Now, with the substitution $t=1/x^2$, you get
$$
lim_{tto0^+}frac{log(1+sinh t)}{t}
$$
answered Jan 3 at 21:53
egreg
179k1484201
179k1484201
add a comment |
add a comment |
This limit becomes obvious if you recall that $sinh x sim x$ as $x to 0$ (just as regular $sin$ is). So substitute $t = {1over x}$:
$$
lim_{tto 0}(1+sinh(t^2))^{1over t^2} sim lim_{tto 0}(1+t^2)^{1over t^2}
$$
RHS is a very famous limit.
add a comment |
This limit becomes obvious if you recall that $sinh x sim x$ as $x to 0$ (just as regular $sin$ is). So substitute $t = {1over x}$:
$$
lim_{tto 0}(1+sinh(t^2))^{1over t^2} sim lim_{tto 0}(1+t^2)^{1over t^2}
$$
RHS is a very famous limit.
add a comment |
This limit becomes obvious if you recall that $sinh x sim x$ as $x to 0$ (just as regular $sin$ is). So substitute $t = {1over x}$:
$$
lim_{tto 0}(1+sinh(t^2))^{1over t^2} sim lim_{tto 0}(1+t^2)^{1over t^2}
$$
RHS is a very famous limit.
This limit becomes obvious if you recall that $sinh x sim x$ as $x to 0$ (just as regular $sin$ is). So substitute $t = {1over x}$:
$$
lim_{tto 0}(1+sinh(t^2))^{1over t^2} sim lim_{tto 0}(1+t^2)^{1over t^2}
$$
RHS is a very famous limit.
answered Jan 3 at 22:22
roman
1,98121221
1,98121221
add a comment |
add a comment |
1
What on earth is "sh?" Do you mean $sinh$, the hyperbolic sine function?
– Mark Viola
Jan 3 at 21:38
Yes, sorry, we use sh for sinh
– MilO
Jan 3 at 21:42
Hi! Welcome to MSE. Please use Mathjax wherever possible and provide a little more context as to why this problem is relevant to you and this community
– Sauhard Sharma
Jan 3 at 21:43
You don't want answers that use L'Hopital's rule?
– 0x539
Jan 3 at 21:53