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$\displaystyle\lim_{x\to 0}^{}\sin{x}^{x}$. I tried this $ P =\displaystyle \lim_{x\to 0}(\sin{x})^x$ taking the natural log on both side we get $\log_{e}P= \displaystyle \lim_{x\to 0}\log_{e}(\sin{x})^x$

$=\displaystyle \lim_{x\to 0}x\ln(\sin{x})$

$=\displaystyle \lim_{x\to 0}\frac{\ln(\sin{x})}{1\over x}$

this is $\frac{\infty}{\infty}$

I think we should use L'Hospital's rule, but I can't find the answer.

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  • $\begingroup$ Well what is $\lim_{x \to 0} x^x$? $\endgroup$ – Aldon May 5 '15 at 16:47
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    $\begingroup$ $\;x^x\;$ doesn't exist for $\;x<0\;$ and thus the wanted limit cannot exist. If you want to talk of the right sided then something can be done...perhaps. $\endgroup$ – Timbuc May 5 '15 at 16:48
  • $\begingroup$ If you mean $\lim_{x\to 0}\sin (x^x)$, then $\sin $ being continuous implies $\lim_{x\to 0^+}\sin x^x=\sin\lim_{x\to 0^+} x^x=\sin 1$. But I see that's not what you meant. $\lim_{x\to 0}\sin x^x$ is ambiguous. $\endgroup$ – user26486 May 5 '15 at 16:59
  • $\begingroup$ @user31415 that is not the question $\endgroup$ – RE60K May 5 '15 at 17:00
  • $\begingroup$ @ADG I know. I just pointed out that $\displaystyle\lim_{x\to 0}^{}\sin{x}^{x}$ is ambiguous, though later statements remove ambiguity. $\endgroup$ – user26486 May 5 '15 at 17:01
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For the right-hand side limit:

Suppose $0<x<\pi$. Then: $$x\ln(\sin x)=\frac x{\sin x}\cdot \sin x \ln(\sin x)$$ Now $\dfrac x{\sin x}\to 1$ as $x\to 0$, and $\,u\ln u\to 0$ as $u\to 0_+$, so
$$x\ln(\sin x)\to 0\enspace\text{and}\quad\lim_{x\to 0_+} (\sin x)^x = 1$$

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$$\lim_{x\to0}(\sin x)^x=\lim_{x\to0}e^{x\ln(\sin x)}=\lim_{x\to0}e^{\displaystyle \frac{\ln(\sin x)}{1/x}}\stackrel{L'H}=\lim_{x\to0}e^{\displaystyle \frac{\cot x}{-1/x^2}}=\lim_{x\to0}e^{\displaystyle -x\times\frac{x}{\tan x}}=e^{0\times1}=1$$

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  • $\begingroup$ This can't be right as $\;x<0\implies (\sin x)^x\;$ isn't defined (for example, try with $\;x=-\frac1{20}\;$ ...) $\endgroup$ – Timbuc May 5 '15 at 17:00
  • $\begingroup$ @Timbuc well you're having it wrong, sorry but please read about limits again. $\endgroup$ – RE60K May 5 '15 at 17:17
  • $\begingroup$ First, I think the function in the limit is actually $\;\sin\left(x^x\right)\;$ and not what you wrote, but whatever: How can a negative number be raised to a power $\;n/m\;$ , with $\;n\;$ an odd natural and $\;m\;$ and even one? Maybe I'm missing something here... $\endgroup$ – Timbuc May 5 '15 at 17:19
  • $\begingroup$ First, see this and yes you are missing something [I'm just avoiding it ATM so that you find it yourself or else, if you need, I may tell you] $\endgroup$ – RE60K May 5 '15 at 17:22
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    $\begingroup$ Sorry, I can't see your point, and I don't know what ATM stands for. And please be kind to me and tell me exactly what it is I'm missing instead of repeating over and over that I'm missing something. Thanks. $\endgroup$ – Timbuc May 5 '15 at 17:27

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