Alternate way to solve $\lim\limits_{x \to 0} (\sin x) ^x$? My solution:
$$ \lim_{x \to 0} (\sin x)^x = \lim_{x \to 0} e^{(x)(\ln \sin x)} = \exp \left( \lim_{x \to 0} (x) (\ln \sin x) \right)$$
Now we have $\lim_{x \to 0} (x) (\ln\sin x)$.
Now  we  can say that the limit is $0$ as $\ln$, $\sin x$ decreases more slowly than $x$.
Question: Is there any other method to solve the above limit without using the arguments saying one function decreases slower?
 A: I would write $x \ln (\sin x)=x \ln (\frac  {\sin x}  x) +x\ln x$. The first term tends to $(0)(0)=0$ so we only have to find $\lim x \ln  x$. For this we can apply L'Hopital's Rule to $\lim \frac {\ln x} {1/x}$.
A: You can apply Hôpital's rule :
$$e^{\lim_{x \to 0} \frac{1}{\frac{1}{x}} \ln{\left(\sin{\left(x \right)} \right)}} = e^{\lim_{x \to 0} \frac{\frac{d}{dx}\left(\ln{\left(\sin{\left(x \right)} \right)}\right)}{\frac{d}{dx}\left(\frac{1}{x}\right)}}=e^{\lim_{x \to 0}\left(- \frac{x^{2} \cos{\left(x \right)}}{\sin{\left(x \right)}}\right)}$$
Focus on
$$\lim_{x \to 0}\left(- \frac{x^{2} \cos{\left(x \right)}}{\sin{\left(x \right)}}\right)$$ you can apply again the Hôpital's rule having an indeterminate form of type $(0/0)$.
A: A shorter approach using L’Hôpital’s : $$\lim_{x\to 0} x\ln(\sin x) \\ = \lim_{x\to 0} \frac{x}{\sin x} \cdot \sin x \ln(\sin x) \\=\lim_{x\to 0} \sin x \ln(\sin x)$$
Substitute $\sin x= h$, $$=\lim_{h\to 0} h\ln h \overset{\text{L.H.}}= 0$$ and so the original limit is $1$.
A: For $x\in \left[0,\frac{\pi}2\right]$ we have
$$\frac 2{\pi}x \leq \sin x \leq x$$
Now, using the standard limit $\lim_{x\to 0^+}x^x = 1$ you get
$$1 = \lim_{x\to 0^+}\left(\left(\frac 2{\pi}\right)^xx^x\right)\leq \lim_{x\to 0^+}\sin^x x \leq \lim_{x\to 0^+}x^x = 1$$
A: A short solution using equivalence of functions near $0$:
We have $\sin x\sim_0 x$, whence $\: x\ln(\sin x)\sim_0 x\ln x$, which tends to $0$ as $x \to 0$. Therefore
$$\lim_{x\to0}(\sin x)^x=\lim_{x\to0}\mathrm e^{x\ln(\sin x)}=\mathrm e^0=1.$$
A: Just a small variant on Kavi Rama Murthy's answer, together with a comment on the one-sidedness of the limit:
Note,
$$(\sin x)^x=\left(\sin x\over x\right)^xx^x$$
We have
$$\lim_{x\to0}\left(\sin x\over x\right)^x=1^0=1$$
and
$$\lim_{x\to0^+}x^x=1$$
(from the easy L'Hopital for $x\ln x={\ln x\over1/x}$). Therefore
$$\lim_{x\to0^+}(\sin x)^x=1$$
Remark: I've specified the limit from the right, $x\to0^+$, since $x^x$ and $(\sin x)^x$ are undefined (as real-valued functions) for (small) negative values of $x$. The function $\left(\sin x\over x\right)^x$ is defined for negative values of $x$ near $0$, so it's OK to consider its limit as $x\to0$ from both sides.
