# How to find limit of $\lim_{x\to 0}\sin x^x$?

$\displaystyle\lim_{x\to 0}^{}\sin{x}^{x}$. 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.

• Well what is $\lim_{x \to 0} x^x$? – Aldon May 5 '15 at 16:47
• $\;x^x\;$ 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. – Timbuc May 5 '15 at 16:48
• 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. – user26486 May 5 '15 at 16:59
• @user31415 that is not the question – RE60K May 5 '15 at 17:00
• @ADG I know. I just pointed out that $\displaystyle\lim_{x\to 0}^{}\sin{x}^{x}$ is ambiguous, though later statements remove ambiguity. – user26486 May 5 '15 at 17:01

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
$$x\ln(\sin x)\to 0\enspace\text{and}\quad\lim_{x\to 0_+} (\sin x)^x = 1$$
$$\lim_{x\to0}(\sin x)^x=\lim_{x\to0}e^{x\ln(\sin x)}=\lim_{x\to0}e^{\displaystyle \frac{\ln(\sin x)}{1/x}}\stackrel{L'H}=\lim_{x\to0}e^{\displaystyle \frac{\cot x}{-1/x^2}}=\lim_{x\to0}e^{\displaystyle -x\times\frac{x}{\tan x}}=e^{0\times1}=1$$
• This can't be right as $\;x<0\implies (\sin x)^x\;$ isn't defined (for example, try with $\;x=-\frac1{20}\;$ ...) – Timbuc May 5 '15 at 17:00
• 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... – Timbuc May 5 '15 at 17:19