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I'm stuck with finding the following limit:

$$\lim_{x\to0} \frac{1}{x}\int_0^{2x} (\sin t)^{t} dt $$

I guess that it is necessary to find the integral first. Having transformed the integrand $(\sin t)^{t} = e^{t \ln (\sin t)}$, I have tried integration by parts and substitution, but it seems to get only worse. Any help would be great!

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    $\begingroup$ use the rules of L'Hospital, the result should be $2$ $\endgroup$ – Dr. Sonnhard Graubner Aug 10 '16 at 15:13
  • $\begingroup$ Use the Taylor expansion of $\sin(x)$. Also note that this is 2 times the derivative of the integrand $\endgroup$ – Oles Wohnzimmer Aug 10 '16 at 15:14
  • $\begingroup$ @Dr.SonnhardGraubner better than most of the answers here. I would post this. $\endgroup$ – Brevan Ellefsen Aug 10 '16 at 15:40
  • $\begingroup$ @Dr.SonnhardGraubner could you please explain how you get 2 in the end? Is it correct that I should use the fundamental theorem of calculus? Isn't the resulting limit $lim_{x\to0} \sin^{2x} (2x) $? $\endgroup$ – alexpetnet Aug 10 '16 at 16:23
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    $\begingroup$ @alexpetnet almost! Remember to use the chain rule, so you actually get $2\sin^{2x}(2x)$ $\endgroup$ – Brevan Ellefsen Aug 10 '16 at 16:38
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We have, from the mean value theorem, that exists some $c_{x}\in\left(0,2x\right) $ such that $$\int_{0}^{2x}\sin^{t}\left(t\right)dt=\sin^{c_{x}}\left(c_{x}\right)\int_{0}^{2x}1dt=2x\sin^{c_{x}}\left(c_{x}\right) $$ and since $$\lim_{x\rightarrow0}\sin^{x}\left(x\right)=1 $$ we have $$\lim_{x\rightarrow0}\sin^{c_{x}}\left(c_{x}\right)=1 $$ hence $$\lim_{x\rightarrow0}\frac{1}{x}\int_{0}^{2x}\sin^{t}\left(t\right)dt=\color{red}{2}.$$

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  • $\begingroup$ Thank you! The idea is clear to me. However, I'd be more thankful if you clarified why it is justified to replace $\lim_{x\to0} \int_{0}^{2x}\sin^{t}\left(t\right)dt$ with $\lim_{c_{x}\to0} 2x\sin^{c_{x}}\left(c_{x}\right)$? $\endgroup$ – alexpetnet Aug 10 '16 at 17:55
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    $\begingroup$ @alexpetnet $c_{x}$ is a (unknown) point in $(0,2x)$. If you take the limit at $x\rightarrow0$ we have necessary that $c_{x}\rightarrow0$ by squeeze theorem. So we have done, since $$\lim_{x\rightarrow0}\sin^{f(x)}(f(x))=1$$ for all $f(x)$ such that $f(x)\rightarrow0$ if $x\rightarrow0$. $\endgroup$ – Marco Cantarini Aug 11 '16 at 6:48
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Hint: if we define $f(t) = \sin(t)^t$ with $f(0) = \lim_{t\to 0} \sin(t)^t = 1$, then for small $x$, $$\int_0^{2x}\sin(t)^tdt = \int_0^{2x} f(t) dt\approx 2xf(0) = 2x$$ This would work with any continuous function.

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