Compute $lim_{xto0} (sin x)^x$ using L'Hospital's rule [closed]
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Help me to compute this limit with L'Hospital's rule.
$$lim_{xto 0} (sin x)^x$$
Thanks in advance.
limits
edited Nov 22 at 11:48
Jean-Claude Arbaut
14.7k63363
asked Nov 22 at 11:40
Rumman Bin Rayhan Rijvi
71
closed as off-topic by amWhy, José Carlos Santos, Jean-Claude Arbaut, Nosrati, Hans Lundmark Nov 22 at 20:22
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Jean-Claude Arbaut, Nosrati, Hans Lundmark
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
-1
down vote
favorite
Help me to compute this limit with L'Hospital's rule.
$$lim_{xto 0} (sin x)^x$$
Thanks in advance.
limits
edited Nov 22 at 11:48
Jean-Claude Arbaut
14.7k63363
asked Nov 22 at 11:40
Rumman Bin Rayhan Rijvi
71
closed as off-topic by amWhy, José Carlos Santos, Jean-Claude Arbaut, Nosrati, Hans Lundmark Nov 22 at 20:22
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Jean-Claude Arbaut, Nosrati, Hans Lundmark
If this question can be reworded to fit the rules in the help center, please edit the question.
Does the limit exist?
– Akash Roy
Nov 22 at 11:42
H Hospitals law limits need to solve this math
– Rumman Bin Rayhan Rijvi
Nov 22 at 11:43
4
No need to comment on someone's grammar mistakes, especially when English are not they're first language.
– Rebellos
Nov 22 at 11:47
L. hospitals law
– Rumman Bin Rayhan Rijvi
Nov 22 at 11:49
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Help me to compute this limit with L'Hospital's rule.
$$lim_{xto 0} (sin x)^x$$
Thanks in advance.
limits
edited Nov 22 at 11:48
Jean-Claude Arbaut
14.7k63363
asked Nov 22 at 11:40
Rumman Bin Rayhan Rijvi
71
Help me to compute this limit with L'Hospital's rule.
$$lim_{xto 0} (sin x)^x$$
Thanks in advance.
limits
limits
edited Nov 22 at 11:48
Jean-Claude Arbaut
14.7k63363
asked Nov 22 at 11:40
Rumman Bin Rayhan Rijvi
71
edited Nov 22 at 11:48
Jean-Claude Arbaut
14.7k63363
asked Nov 22 at 11:40
Rumman Bin Rayhan Rijvi
71
edited Nov 22 at 11:48
Jean-Claude Arbaut
14.7k63363
edited Nov 22 at 11:48
Jean-Claude Arbaut
14.7k63363
edited Nov 22 at 11:48
Jean-Claude Arbaut
14.7k63363
14.7k63363
asked Nov 22 at 11:40
Rumman Bin Rayhan Rijvi
71
asked Nov 22 at 11:40
Rumman Bin Rayhan Rijvi
71
asked Nov 22 at 11:40
Rumman Bin Rayhan Rijvi
71
71
closed as off-topic by amWhy, José Carlos Santos, Jean-Claude Arbaut, Nosrati, Hans Lundmark Nov 22 at 20:22
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Jean-Claude Arbaut, Nosrati, Hans Lundmark
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, José Carlos Santos, Jean-Claude Arbaut, Nosrati, Hans Lundmark Nov 22 at 20:22
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Jean-Claude Arbaut, Nosrati, Hans Lundmark
If this question can be reworded to fit the rules in the help center, please edit the question.
Does the limit exist?
– Akash Roy
Nov 22 at 11:42
H Hospitals law limits need to solve this math
– Rumman Bin Rayhan Rijvi
Nov 22 at 11:43
4
No need to comment on someone's grammar mistakes, especially when English are not they're first language.
– Rebellos
Nov 22 at 11:47
L. hospitals law
– Rumman Bin Rayhan Rijvi
Nov 22 at 11:49
add a comment |
Does the limit exist?
– Akash Roy
Nov 22 at 11:42
H Hospitals law limits need to solve this math
– Rumman Bin Rayhan Rijvi
Nov 22 at 11:43
4
No need to comment on someone's grammar mistakes, especially when English are not they're first language.
– Rebellos
Nov 22 at 11:47
L. hospitals law
– Rumman Bin Rayhan Rijvi
Nov 22 at 11:49
Does the limit exist?
– Akash Roy
Nov 22 at 11:42
Does the limit exist?
– Akash Roy
Nov 22 at 11:42
H Hospitals law limits need to solve this math
– Rumman Bin Rayhan Rijvi
Nov 22 at 11:43
H Hospitals law limits need to solve this math
– Rumman Bin Rayhan Rijvi
Nov 22 at 11:43
4
4
No need to comment on someone's grammar mistakes, especially when English are not they're first language.
– Rebellos
Nov 22 at 11:47
No need to comment on someone's grammar mistakes, especially when English are not they're first language.
– Rebellos
Nov 22 at 11:47
L. hospitals law
– Rumman Bin Rayhan Rijvi
Nov 22 at 11:49
L. hospitals law
– Rumman Bin Rayhan Rijvi
Nov 22 at 11:49
add a comment |
2 Answers
2
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oldest
votes
up vote
2
down vote
HINT
We need $x>0$, then use
$$ (sin x)^x=e^{x log sin x}=e^{frac{log sin x}{frac1x}}$$
answered Nov 22 at 11:47
gimusi
91.4k74495
add a comment |
up vote
0
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Use L'Hospital's rule to complete the standard exercise of finding that
$$ lim_{xto0}frac{sin x}{x}=1.$$
So we have that, as $xto0$, $sin xsim x$. Now the limit you have becomes
$$L=lim_{xto0}(sin x)^x = lim_{xto 0}x^x = lim_{xto0}e^{xln x}.$$
Now, take the log of this limit and find $ln L$. Once you've done that, just exponentiate this value to find $L$.
answered Nov 22 at 12:22
YiFan
2,0841419
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
HINT
We need $x>0$, then use
$$ (sin x)^x=e^{x log sin x}=e^{frac{log sin x}{frac1x}}$$
answered Nov 22 at 11:47
gimusi
91.4k74495
add a comment |
up vote
2
down vote
HINT
We need $x>0$, then use
$$ (sin x)^x=e^{x log sin x}=e^{frac{log sin x}{frac1x}}$$
answered Nov 22 at 11:47
gimusi
91.4k74495
add a comment |
up vote
2
down vote
up vote
2
down vote
HINT
We need $x>0$, then use
$$ (sin x)^x=e^{x log sin x}=e^{frac{log sin x}{frac1x}}$$
answered Nov 22 at 11:47
gimusi
91.4k74495
HINT
We need $x>0$, then use
$$ (sin x)^x=e^{x log sin x}=e^{frac{log sin x}{frac1x}}$$
answered Nov 22 at 11:47
gimusi
91.4k74495
answered Nov 22 at 11:47
gimusi
91.4k74495
answered Nov 22 at 11:47
gimusi
91.4k74495
answered Nov 22 at 11:47
gimusi
91.4k74495
91.4k74495
add a comment |
add a comment |
up vote
0
down vote
Use L'Hospital's rule to complete the standard exercise of finding that
$$ lim_{xto0}frac{sin x}{x}=1.$$
So we have that, as $xto0$, $sin xsim x$. Now the limit you have becomes
$$L=lim_{xto0}(sin x)^x = lim_{xto 0}x^x = lim_{xto0}e^{xln x}.$$
Now, take the log of this limit and find $ln L$. Once you've done that, just exponentiate this value to find $L$.
answered Nov 22 at 12:22
YiFan
2,0841419
add a comment |
up vote
0
down vote
Use L'Hospital's rule to complete the standard exercise of finding that
$$ lim_{xto0}frac{sin x}{x}=1.$$
So we have that, as $xto0$, $sin xsim x$. Now the limit you have becomes
$$L=lim_{xto0}(sin x)^x = lim_{xto 0}x^x = lim_{xto0}e^{xln x}.$$
Now, take the log of this limit and find $ln L$. Once you've done that, just exponentiate this value to find $L$.
answered Nov 22 at 12:22
YiFan
2,0841419
add a comment |
up vote
0
down vote
up vote
0
down vote
Use L'Hospital's rule to complete the standard exercise of finding that
$$ lim_{xto0}frac{sin x}{x}=1.$$
So we have that, as $xto0$, $sin xsim x$. Now the limit you have becomes
$$L=lim_{xto0}(sin x)^x = lim_{xto 0}x^x = lim_{xto0}e^{xln x}.$$
Now, take the log of this limit and find $ln L$. Once you've done that, just exponentiate this value to find $L$.
answered Nov 22 at 12:22
YiFan
2,0841419
Use L'Hospital's rule to complete the standard exercise of finding that
$$ lim_{xto0}frac{sin x}{x}=1.$$
So we have that, as $xto0$, $sin xsim x$. Now the limit you have becomes
$$L=lim_{xto0}(sin x)^x = lim_{xto 0}x^x = lim_{xto0}e^{xln x}.$$
Now, take the log of this limit and find $ln L$. Once you've done that, just exponentiate this value to find $L$.
answered Nov 22 at 12:22
YiFan
2,0841419
answered Nov 22 at 12:22
YiFan
2,0841419
answered Nov 22 at 12:22
YiFan
2,0841419
answered Nov 22 at 12:22
YiFan
2,0841419
2,0841419
add a comment |
add a comment |
Does the limit exist?
– Akash Roy
Nov 22 at 11:42
H Hospitals law limits need to solve this math
– Rumman Bin Rayhan Rijvi
Nov 22 at 11:43
4
No need to comment on someone's grammar mistakes, especially when English are not they're first language.
– Rebellos
Nov 22 at 11:47
L. hospitals law
– Rumman Bin Rayhan Rijvi
Nov 22 at 11:49