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Find limit $lim_{xrightarrow 0}frac{2^{sin(x)}-1}{x} = 0 $
0
$begingroup$
can someone provide me with some hint how to evaluate this limit?
$$lim_{xrightarrow 0}frac{2^{sin(x)}-1}{x} = 0 $$
Unfortunately, I can't use l'hopital's rule
I was thinking about something like that:
$$lim_{xrightarrow 0}frac{2^{sin(x)}-1}{x} =\lim_{xrightarrow 0}frac{ln(e^{2^{sin(x)}})-1}{ln(e^x)} $$ but there I don't see how to continue this way of thinking (of course if it is correct)
real-analysis limits-without-lhopital
asked Jan 20 at 10:48
VirtualUserVirtualUser
996115
$endgroup$
add a comment |
0
$begingroup$
can someone provide me with some hint how to evaluate this limit?
$$lim_{xrightarrow 0}frac{2^{sin(x)}-1}{x} = 0 $$
Unfortunately, I can't use l'hopital's rule
I was thinking about something like that:
$$lim_{xrightarrow 0}frac{2^{sin(x)}-1}{x} =\lim_{xrightarrow 0}frac{ln(e^{2^{sin(x)}})-1}{ln(e^x)} $$ but there I don't see how to continue this way of thinking (of course if it is correct)
real-analysis limits-without-lhopital
asked Jan 20 at 10:48
VirtualUserVirtualUser
996115
$endgroup$
3
$begingroup$
The limit is precisely the definition of $f'(0)$ where $f(x) = 2^{sin(x)}$. This technically doesn't use L'Hopital's rule! ;-)
$endgroup$
– Theo Bendit
Jan 20 at 10:58
$begingroup$
"Unfortunately, I can't use l'hopital's rule" Correction: Fortunately, you cannot use LH.
$endgroup$
– Did
Jan 20 at 11:10
$begingroup$
Just combine $lim_{zto 0}frac{a^z-1}{z}=log a$ with $lim_{wto 0}frac{sin w}{w}=1$ to get $color{red}{log 2}neq 0$.
$endgroup$
– Jack D'Aurizio
Jan 20 at 18:44
add a comment |
0
0
0
$begingroup$
can someone provide me with some hint how to evaluate this limit?
$$lim_{xrightarrow 0}frac{2^{sin(x)}-1}{x} = 0 $$
Unfortunately, I can't use l'hopital's rule
I was thinking about something like that:
$$lim_{xrightarrow 0}frac{2^{sin(x)}-1}{x} =\lim_{xrightarrow 0}frac{ln(e^{2^{sin(x)}})-1}{ln(e^x)} $$ but there I don't see how to continue this way of thinking (of course if it is correct)
real-analysis limits-without-lhopital
asked Jan 20 at 10:48
VirtualUserVirtualUser
996115
$endgroup$
can someone provide me with some hint how to evaluate this limit?
$$lim_{xrightarrow 0}frac{2^{sin(x)}-1}{x} = 0 $$
Unfortunately, I can't use l'hopital's rule
I was thinking about something like that:
$$lim_{xrightarrow 0}frac{2^{sin(x)}-1}{x} =\lim_{xrightarrow 0}frac{ln(e^{2^{sin(x)}})-1}{ln(e^x)} $$ but there I don't see how to continue this way of thinking (of course if it is correct)
real-analysis limits-without-lhopital
real-analysis limits-without-lhopital
asked Jan 20 at 10:48
VirtualUserVirtualUser
996115
asked Jan 20 at 10:48
VirtualUserVirtualUser
996115
asked Jan 20 at 10:48
VirtualUserVirtualUser
996115
asked Jan 20 at 10:48
VirtualUserVirtualUser
996115
asked Jan 20 at 10:48
VirtualUserVirtualUser
996115
996115
3
$begingroup$
The limit is precisely the definition of $f'(0)$ where $f(x) = 2^{sin(x)}$. This technically doesn't use L'Hopital's rule! ;-)
$endgroup$
– Theo Bendit
Jan 20 at 10:58
$begingroup$
"Unfortunately, I can't use l'hopital's rule" Correction: Fortunately, you cannot use LH.
$endgroup$
– Did
Jan 20 at 11:10
$begingroup$
Just combine $lim_{zto 0}frac{a^z-1}{z}=log a$ with $lim_{wto 0}frac{sin w}{w}=1$ to get $color{red}{log 2}neq 0$.
$endgroup$
– Jack D'Aurizio
Jan 20 at 18:44
add a comment |
3
$begingroup$
The limit is precisely the definition of $f'(0)$ where $f(x) = 2^{sin(x)}$. This technically doesn't use L'Hopital's rule! ;-)
$endgroup$
– Theo Bendit
Jan 20 at 10:58
$begingroup$
"Unfortunately, I can't use l'hopital's rule" Correction: Fortunately, you cannot use LH.
$endgroup$
– Did
Jan 20 at 11:10
$begingroup$
Just combine $lim_{zto 0}frac{a^z-1}{z}=log a$ with $lim_{wto 0}frac{sin w}{w}=1$ to get $color{red}{log 2}neq 0$.
$endgroup$
– Jack D'Aurizio
Jan 20 at 18:44
3
3
$begingroup$
The limit is precisely the definition of $f'(0)$ where $f(x) = 2^{sin(x)}$. This technically doesn't use L'Hopital's rule! ;-)
$endgroup$
– Theo Bendit
Jan 20 at 10:58
$begingroup$
The limit is precisely the definition of $f'(0)$ where $f(x) = 2^{sin(x)}$. This technically doesn't use L'Hopital's rule! ;-)
$endgroup$
– Theo Bendit
Jan 20 at 10:58
$begingroup$
"Unfortunately, I can't use l'hopital's rule" Correction: Fortunately, you cannot use LH.
$endgroup$
– Did
Jan 20 at 11:10
$begingroup$
"Unfortunately, I can't use l'hopital's rule" Correction: Fortunately, you cannot use LH.
$endgroup$
– Did
Jan 20 at 11:10
$begingroup$
Just combine $lim_{zto 0}frac{a^z-1}{z}=log a$ with $lim_{wto 0}frac{sin w}{w}=1$ to get $color{red}{log 2}neq 0$.
$endgroup$
– Jack D'Aurizio
Jan 20 at 18:44
$begingroup$
Just combine $lim_{zto 0}frac{a^z-1}{z}=log a$ with $lim_{wto 0}frac{sin w}{w}=1$ to get $color{red}{log 2}neq 0$.
$endgroup$
– Jack D'Aurizio
Jan 20 at 18:44
add a comment |
1 Answer
1
active
oldest
votes
2
$begingroup$
Hint:
For $sin xne0$
$$dfrac{2^{sin x}-1}x=dfrac{2^{sin x}-1}{sin x}cdotdfrac{sin x}x$$
$$implieslim_{xto0}dfrac{2^{sin x}-1}x=lim_{xto0}dfrac{2^{sin x}-1}{sin x}cdotlim_{xto0}dfrac{sin x}x$$
answered Jan 20 at 10:49
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
$begingroup$
Ok, second part is a theorem which I know, but first part? It is needed to transform it or you mean about theorem there?
$endgroup$
– VirtualUser
Jan 20 at 10:53
$begingroup$
Why is the limit of the remaining ratio supposed to be easier to determine than the limit of the original ratio?
$endgroup$
– Did
Jan 20 at 11:11
2
$begingroup$
The former limit $limlimits_{xto 0}frac{2^{sin x}-1}{sin x}$ can be solved by substituting $t=sin x$. You then get $limlimits_{tto 0}frac{2^t-1}{t}$. Can you proceed?
$endgroup$
– Yuval Gat
Jan 20 at 11:23
$begingroup$
@VirtualUser, Use math.stackexchange.com/questions/942565/…
$endgroup$
– lab bhattacharjee
Jan 20 at 12:45
add a comment |
Your Answer
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1 Answer
1
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oldest
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1 Answer
1
active
oldest
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votes
2
$begingroup$
Hint:
For $sin xne0$
$$dfrac{2^{sin x}-1}x=dfrac{2^{sin x}-1}{sin x}cdotdfrac{sin x}x$$
$$implieslim_{xto0}dfrac{2^{sin x}-1}x=lim_{xto0}dfrac{2^{sin x}-1}{sin x}cdotlim_{xto0}dfrac{sin x}x$$
answered Jan 20 at 10:49
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
$begingroup$
Ok, second part is a theorem which I know, but first part? It is needed to transform it or you mean about theorem there?
$endgroup$
– VirtualUser
Jan 20 at 10:53
$begingroup$
Why is the limit of the remaining ratio supposed to be easier to determine than the limit of the original ratio?
$endgroup$
– Did
Jan 20 at 11:11
2
$begingroup$
The former limit $limlimits_{xto 0}frac{2^{sin x}-1}{sin x}$ can be solved by substituting $t=sin x$. You then get $limlimits_{tto 0}frac{2^t-1}{t}$. Can you proceed?
$endgroup$
– Yuval Gat
Jan 20 at 11:23
$begingroup$
@VirtualUser, Use math.stackexchange.com/questions/942565/…
$endgroup$
– lab bhattacharjee
Jan 20 at 12:45
add a comment |
2
$begingroup$
Hint:
For $sin xne0$
$$dfrac{2^{sin x}-1}x=dfrac{2^{sin x}-1}{sin x}cdotdfrac{sin x}x$$
$$implieslim_{xto0}dfrac{2^{sin x}-1}x=lim_{xto0}dfrac{2^{sin x}-1}{sin x}cdotlim_{xto0}dfrac{sin x}x$$
answered Jan 20 at 10:49
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
$begingroup$
Ok, second part is a theorem which I know, but first part? It is needed to transform it or you mean about theorem there?
$endgroup$
– VirtualUser
Jan 20 at 10:53
$begingroup$
Why is the limit of the remaining ratio supposed to be easier to determine than the limit of the original ratio?
$endgroup$
– Did
Jan 20 at 11:11
2
$begingroup$
The former limit $limlimits_{xto 0}frac{2^{sin x}-1}{sin x}$ can be solved by substituting $t=sin x$. You then get $limlimits_{tto 0}frac{2^t-1}{t}$. Can you proceed?
$endgroup$
– Yuval Gat
Jan 20 at 11:23
$begingroup$
@VirtualUser, Use math.stackexchange.com/questions/942565/…
$endgroup$
– lab bhattacharjee
Jan 20 at 12:45
add a comment |
2
2
2
$begingroup$
Hint:
For $sin xne0$
$$dfrac{2^{sin x}-1}x=dfrac{2^{sin x}-1}{sin x}cdotdfrac{sin x}x$$
$$implieslim_{xto0}dfrac{2^{sin x}-1}x=lim_{xto0}dfrac{2^{sin x}-1}{sin x}cdotlim_{xto0}dfrac{sin x}x$$
answered Jan 20 at 10:49
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
Hint:
For $sin xne0$
$$dfrac{2^{sin x}-1}x=dfrac{2^{sin x}-1}{sin x}cdotdfrac{sin x}x$$
$$implieslim_{xto0}dfrac{2^{sin x}-1}x=lim_{xto0}dfrac{2^{sin x}-1}{sin x}cdotlim_{xto0}dfrac{sin x}x$$
answered Jan 20 at 10:49
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 20 at 10:49
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 20 at 10:49
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 20 at 10:49
lab bhattacharjeelab bhattacharjee
226k15157275
226k15157275
$begingroup$
Ok, second part is a theorem which I know, but first part? It is needed to transform it or you mean about theorem there?
$endgroup$
– VirtualUser
Jan 20 at 10:53
$begingroup$
Why is the limit of the remaining ratio supposed to be easier to determine than the limit of the original ratio?
$endgroup$
– Did
Jan 20 at 11:11
2
$begingroup$
The former limit $limlimits_{xto 0}frac{2^{sin x}-1}{sin x}$ can be solved by substituting $t=sin x$. You then get $limlimits_{tto 0}frac{2^t-1}{t}$. Can you proceed?
$endgroup$
– Yuval Gat
Jan 20 at 11:23
$begingroup$
@VirtualUser, Use math.stackexchange.com/questions/942565/…
$endgroup$
– lab bhattacharjee
Jan 20 at 12:45
add a comment |
$begingroup$
Ok, second part is a theorem which I know, but first part? It is needed to transform it or you mean about theorem there?
$endgroup$
– VirtualUser
Jan 20 at 10:53
$begingroup$
Why is the limit of the remaining ratio supposed to be easier to determine than the limit of the original ratio?
$endgroup$
– Did
Jan 20 at 11:11
2
$begingroup$
The former limit $limlimits_{xto 0}frac{2^{sin x}-1}{sin x}$ can be solved by substituting $t=sin x$. You then get $limlimits_{tto 0}frac{2^t-1}{t}$. Can you proceed?
$endgroup$
– Yuval Gat
Jan 20 at 11:23
$begingroup$
@VirtualUser, Use math.stackexchange.com/questions/942565/…
$endgroup$
– lab bhattacharjee
Jan 20 at 12:45
$begingroup$
Ok, second part is a theorem which I know, but first part? It is needed to transform it or you mean about theorem there?
$endgroup$
– VirtualUser
Jan 20 at 10:53
$begingroup$
Ok, second part is a theorem which I know, but first part? It is needed to transform it or you mean about theorem there?
$endgroup$
– VirtualUser
Jan 20 at 10:53
$begingroup$
Why is the limit of the remaining ratio supposed to be easier to determine than the limit of the original ratio?
$endgroup$
– Did
Jan 20 at 11:11
$begingroup$
Why is the limit of the remaining ratio supposed to be easier to determine than the limit of the original ratio?
$endgroup$
– Did
Jan 20 at 11:11
2
2
$begingroup$
The former limit $limlimits_{xto 0}frac{2^{sin x}-1}{sin x}$ can be solved by substituting $t=sin x$. You then get $limlimits_{tto 0}frac{2^t-1}{t}$. Can you proceed?
$endgroup$
– Yuval Gat
Jan 20 at 11:23
$begingroup$
The former limit $limlimits_{xto 0}frac{2^{sin x}-1}{sin x}$ can be solved by substituting $t=sin x$. You then get $limlimits_{tto 0}frac{2^t-1}{t}$. Can you proceed?
$endgroup$
– Yuval Gat
Jan 20 at 11:23
$begingroup$
@VirtualUser, Use math.stackexchange.com/questions/942565/…
$endgroup$
– lab bhattacharjee
Jan 20 at 12:45
$begingroup$
@VirtualUser, Use math.stackexchange.com/questions/942565/…
$endgroup$
– lab bhattacharjee
Jan 20 at 12:45
add a comment |
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3
$begingroup$
The limit is precisely the definition of $f'(0)$ where $f(x) = 2^{sin(x)}$. This technically doesn't use L'Hopital's rule! ;-)
$endgroup$
– Theo Bendit
Jan 20 at 10:58
$begingroup$
"Unfortunately, I can't use l'hopital's rule" Correction: Fortunately, you cannot use LH.
$endgroup$
– Did
Jan 20 at 11:10
$begingroup$
Just combine $lim_{zto 0}frac{a^z-1}{z}=log a$ with $lim_{wto 0}frac{sin w}{w}=1$ to get $color{red}{log 2}neq 0$.
$endgroup$
– Jack D'Aurizio
Jan 20 at 18:44