Help with a limit (without L'Hopital): $lim_{xto 0^+} x^{sin x}$ [closed]
What is the limit?
$$lim_{xto 0^+} x^{sin x }$$
calculus limits trigonometry limits-without-lhopital
New contributor
closed as off-topic by Henrik, Eevee Trainer, Shaun, Davide Giraudo, Saad Dec 27 '18 at 12:08
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What is the limit?
$$lim_{xto 0^+} x^{sin x }$$
calculus limits trigonometry limits-without-lhopital
New contributor
closed as off-topic by Henrik, Eevee Trainer, Shaun, Davide Giraudo, Saad Dec 27 '18 at 12:08
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henrik, Eevee Trainer, Shaun, Davide Giraudo, Saad
If this question can be reworded to fit the rules in the help center, please edit the question.
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Dec 27 '18 at 9:28
Here's a MathJax tutorial :)
– Shaun
Dec 27 '18 at 9:41
add a comment |
What is the limit?
$$lim_{xto 0^+} x^{sin x }$$
calculus limits trigonometry limits-without-lhopital
New contributor
What is the limit?
$$lim_{xto 0^+} x^{sin x }$$
calculus limits trigonometry limits-without-lhopital
calculus limits trigonometry limits-without-lhopital
New contributor
New contributor
edited Dec 27 '18 at 10:49
Blue
47.6k870151
47.6k870151
New contributor
asked Dec 27 '18 at 9:24
Amit Yeffet
12
12
New contributor
New contributor
closed as off-topic by Henrik, Eevee Trainer, Shaun, Davide Giraudo, Saad Dec 27 '18 at 12:08
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henrik, Eevee Trainer, Shaun, Davide Giraudo, Saad
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Henrik, Eevee Trainer, Shaun, Davide Giraudo, Saad Dec 27 '18 at 12:08
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henrik, Eevee Trainer, Shaun, Davide Giraudo, Saad
If this question can be reworded to fit the rules in the help center, please edit the question.
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Dec 27 '18 at 9:28
Here's a MathJax tutorial :)
– Shaun
Dec 27 '18 at 9:41
add a comment |
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Dec 27 '18 at 9:28
Here's a MathJax tutorial :)
– Shaun
Dec 27 '18 at 9:41
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Dec 27 '18 at 9:28
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Dec 27 '18 at 9:28
Here's a MathJax tutorial :)
– Shaun
Dec 27 '18 at 9:41
Here's a MathJax tutorial :)
– Shaun
Dec 27 '18 at 9:41
add a comment |
2 Answers
2
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oldest
votes
Rewrite as
$$
x^{sin x}=e^{sin xlog x}=e^{frac{sin x}{x}(xlog x)}
$$
and using $sin x/xto 1$ and $xlog xto 0$, the limit is $e^0=1$.
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Hint: Use the expansion for $sin x$.
$$sin x = x-frac{x^3}{3!}+frac{x^5}{5!}-…$$
Apply this to
$$x^{sin x}$$
and let $x to 0^+$. Also, note that
$$lim_{ x to 0^+} x^x = 1$$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Rewrite as
$$
x^{sin x}=e^{sin xlog x}=e^{frac{sin x}{x}(xlog x)}
$$
and using $sin x/xto 1$ and $xlog xto 0$, the limit is $e^0=1$.
add a comment |
Rewrite as
$$
x^{sin x}=e^{sin xlog x}=e^{frac{sin x}{x}(xlog x)}
$$
and using $sin x/xto 1$ and $xlog xto 0$, the limit is $e^0=1$.
add a comment |
Rewrite as
$$
x^{sin x}=e^{sin xlog x}=e^{frac{sin x}{x}(xlog x)}
$$
and using $sin x/xto 1$ and $xlog xto 0$, the limit is $e^0=1$.
Rewrite as
$$
x^{sin x}=e^{sin xlog x}=e^{frac{sin x}{x}(xlog x)}
$$
and using $sin x/xto 1$ and $xlog xto 0$, the limit is $e^0=1$.
answered Dec 27 '18 at 9:28
Pierpaolo Vivo
5,3962624
5,3962624
add a comment |
add a comment |
Hint: Use the expansion for $sin x$.
$$sin x = x-frac{x^3}{3!}+frac{x^5}{5!}-…$$
Apply this to
$$x^{sin x}$$
and let $x to 0^+$. Also, note that
$$lim_{ x to 0^+} x^x = 1$$
add a comment |
Hint: Use the expansion for $sin x$.
$$sin x = x-frac{x^3}{3!}+frac{x^5}{5!}-…$$
Apply this to
$$x^{sin x}$$
and let $x to 0^+$. Also, note that
$$lim_{ x to 0^+} x^x = 1$$
add a comment |
Hint: Use the expansion for $sin x$.
$$sin x = x-frac{x^3}{3!}+frac{x^5}{5!}-…$$
Apply this to
$$x^{sin x}$$
and let $x to 0^+$. Also, note that
$$lim_{ x to 0^+} x^x = 1$$
Hint: Use the expansion for $sin x$.
$$sin x = x-frac{x^3}{3!}+frac{x^5}{5!}-…$$
Apply this to
$$x^{sin x}$$
and let $x to 0^+$. Also, note that
$$lim_{ x to 0^+} x^x = 1$$
edited Dec 27 '18 at 9:33
answered Dec 27 '18 at 9:28
KM101
5,0391423
5,0391423
add a comment |
add a comment |
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Dec 27 '18 at 9:28
Here's a MathJax tutorial :)
– Shaun
Dec 27 '18 at 9:41