Limit using l'Hopital's rule with logaritmus
What is $lim_{xrightarrow0}(1+2sin x)^frac{1}{tan x}$?
Computing $lim_{xrightarrow0}log(1+2sin x)^frac{1}{tan x}=lim_{xrightarrow0}frac{log(1+2sin x)}{tan x}$
Here, where I can use l'Hopital's rule I get:
$lim_{xrightarrow0}frac{log(1+2sin x)}{tan x}=2$
But now my question is:
$lim_{xrightarrow0}(1+2sin x)^frac{1}{tan x}$ = $lim_{xrightarrow0}frac{log(1+2sin x)}{tan x}=2$?
I don't think they are the same, but can I remove the log? Maybe it's easier than I think, but now I don't know how to go on.
real-analysis limits
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What is $lim_{xrightarrow0}(1+2sin x)^frac{1}{tan x}$?
Computing $lim_{xrightarrow0}log(1+2sin x)^frac{1}{tan x}=lim_{xrightarrow0}frac{log(1+2sin x)}{tan x}$
Here, where I can use l'Hopital's rule I get:
$lim_{xrightarrow0}frac{log(1+2sin x)}{tan x}=2$
But now my question is:
$lim_{xrightarrow0}(1+2sin x)^frac{1}{tan x}$ = $lim_{xrightarrow0}frac{log(1+2sin x)}{tan x}=2$?
I don't think they are the same, but can I remove the log? Maybe it's easier than I think, but now I don't know how to go on.
real-analysis limits
add a comment |
What is $lim_{xrightarrow0}(1+2sin x)^frac{1}{tan x}$?
Computing $lim_{xrightarrow0}log(1+2sin x)^frac{1}{tan x}=lim_{xrightarrow0}frac{log(1+2sin x)}{tan x}$
Here, where I can use l'Hopital's rule I get:
$lim_{xrightarrow0}frac{log(1+2sin x)}{tan x}=2$
But now my question is:
$lim_{xrightarrow0}(1+2sin x)^frac{1}{tan x}$ = $lim_{xrightarrow0}frac{log(1+2sin x)}{tan x}=2$?
I don't think they are the same, but can I remove the log? Maybe it's easier than I think, but now I don't know how to go on.
real-analysis limits
What is $lim_{xrightarrow0}(1+2sin x)^frac{1}{tan x}$?
Computing $lim_{xrightarrow0}log(1+2sin x)^frac{1}{tan x}=lim_{xrightarrow0}frac{log(1+2sin x)}{tan x}$
Here, where I can use l'Hopital's rule I get:
$lim_{xrightarrow0}frac{log(1+2sin x)}{tan x}=2$
But now my question is:
$lim_{xrightarrow0}(1+2sin x)^frac{1}{tan x}$ = $lim_{xrightarrow0}frac{log(1+2sin x)}{tan x}=2$?
I don't think they are the same, but can I remove the log? Maybe it's easier than I think, but now I don't know how to go on.
real-analysis limits
real-analysis limits
edited Nov 29 '18 at 22:31
gimusi
1
1
asked Nov 29 '18 at 22:19
DadaDada
7010
7010
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3 Answers
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You mean $$lim_{xto 0}(1+2sin x)^{tfrac{1}{tan x}}=explim_{xto 0}frac{ln (1+2sin x)}{tan x}=e^2.$$
add a comment |
The first equality you write is incorrect. You can evaluate
$$
lim_{xto0}frac{log(1+2sin x)}{tan x}
$$
in order to compute the given limit, but they are not equal; if $l$ is the latter limit, then the one you're looking for is $e^l$.
Now, with l'Hôpital,
$$
lim_{xto0}frac{log(1+2sin x)}{tan x}=
lim_{xto0}frac{dfrac{2cos x}{1+2sin x}}{1+tan^2x}=2
$$
and therefore
$$
lim_{xto0}(1+2sin x)^{1/tan x}=e^2
$$
add a comment |
As an alternative
$$large (1+2 sin x)^frac{1}{tan x}=left[(1+2 sin x)^{frac1{2sin x}}right]^frac{2sin x}{tan x} to e^2$$
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
You mean $$lim_{xto 0}(1+2sin x)^{tfrac{1}{tan x}}=explim_{xto 0}frac{ln (1+2sin x)}{tan x}=e^2.$$
add a comment |
You mean $$lim_{xto 0}(1+2sin x)^{tfrac{1}{tan x}}=explim_{xto 0}frac{ln (1+2sin x)}{tan x}=e^2.$$
add a comment |
You mean $$lim_{xto 0}(1+2sin x)^{tfrac{1}{tan x}}=explim_{xto 0}frac{ln (1+2sin x)}{tan x}=e^2.$$
You mean $$lim_{xto 0}(1+2sin x)^{tfrac{1}{tan x}}=explim_{xto 0}frac{ln (1+2sin x)}{tan x}=e^2.$$
answered Nov 29 '18 at 22:26
J.G.J.G.
23.5k22237
23.5k22237
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The first equality you write is incorrect. You can evaluate
$$
lim_{xto0}frac{log(1+2sin x)}{tan x}
$$
in order to compute the given limit, but they are not equal; if $l$ is the latter limit, then the one you're looking for is $e^l$.
Now, with l'Hôpital,
$$
lim_{xto0}frac{log(1+2sin x)}{tan x}=
lim_{xto0}frac{dfrac{2cos x}{1+2sin x}}{1+tan^2x}=2
$$
and therefore
$$
lim_{xto0}(1+2sin x)^{1/tan x}=e^2
$$
add a comment |
The first equality you write is incorrect. You can evaluate
$$
lim_{xto0}frac{log(1+2sin x)}{tan x}
$$
in order to compute the given limit, but they are not equal; if $l$ is the latter limit, then the one you're looking for is $e^l$.
Now, with l'Hôpital,
$$
lim_{xto0}frac{log(1+2sin x)}{tan x}=
lim_{xto0}frac{dfrac{2cos x}{1+2sin x}}{1+tan^2x}=2
$$
and therefore
$$
lim_{xto0}(1+2sin x)^{1/tan x}=e^2
$$
add a comment |
The first equality you write is incorrect. You can evaluate
$$
lim_{xto0}frac{log(1+2sin x)}{tan x}
$$
in order to compute the given limit, but they are not equal; if $l$ is the latter limit, then the one you're looking for is $e^l$.
Now, with l'Hôpital,
$$
lim_{xto0}frac{log(1+2sin x)}{tan x}=
lim_{xto0}frac{dfrac{2cos x}{1+2sin x}}{1+tan^2x}=2
$$
and therefore
$$
lim_{xto0}(1+2sin x)^{1/tan x}=e^2
$$
The first equality you write is incorrect. You can evaluate
$$
lim_{xto0}frac{log(1+2sin x)}{tan x}
$$
in order to compute the given limit, but they are not equal; if $l$ is the latter limit, then the one you're looking for is $e^l$.
Now, with l'Hôpital,
$$
lim_{xto0}frac{log(1+2sin x)}{tan x}=
lim_{xto0}frac{dfrac{2cos x}{1+2sin x}}{1+tan^2x}=2
$$
and therefore
$$
lim_{xto0}(1+2sin x)^{1/tan x}=e^2
$$
answered Nov 29 '18 at 22:26
egregegreg
179k1485202
179k1485202
add a comment |
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As an alternative
$$large (1+2 sin x)^frac{1}{tan x}=left[(1+2 sin x)^{frac1{2sin x}}right]^frac{2sin x}{tan x} to e^2$$
add a comment |
As an alternative
$$large (1+2 sin x)^frac{1}{tan x}=left[(1+2 sin x)^{frac1{2sin x}}right]^frac{2sin x}{tan x} to e^2$$
add a comment |
As an alternative
$$large (1+2 sin x)^frac{1}{tan x}=left[(1+2 sin x)^{frac1{2sin x}}right]^frac{2sin x}{tan x} to e^2$$
As an alternative
$$large (1+2 sin x)^frac{1}{tan x}=left[(1+2 sin x)^{frac1{2sin x}}right]^frac{2sin x}{tan x} to e^2$$
answered Nov 29 '18 at 22:28
gimusigimusi
1
1
add a comment |
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