Why does $\lim_{x\to 0^+} x^{\sin x}=1$? I tried writing $y=x^{\sin x}$, so $\ln y=\sin x\ln x$. I tried to rewrite the function as $\ln(x)/\csc x$ and apply l'Hopital to the last function, but it's a mess. 
Is there a by hand way to do it?
 A: Your first approach using $\ln$ is good.
So set $$\ln y = \sin x \ln x$$
As $x\to 0^+$, $\sin x \approx x$ therefore: $$\ln y \approx x\ln x \to 0$$ as $x\to 0^+$.
So $y\to 1$
A: Your idea is right. May be you made a mistake in your calculations. Here is what I have
$$
\lim_{x\to0} \frac{\ln x}{1/\sin x} = \lim_{x\to0} \frac{\frac{d}{dx} \ln x}{\frac{d}{dx} (1/\sin x)} = \lim_{x\to0}\frac{1/x}{-\cos x/\sin^2x} =\lim_{x\to0} -\frac{\sin^2x}{x\cos x}$$
The last limit can be written as
$$
\lim_{x\to0} -\frac{\sin^2x}{x\cos x}
=\lim_{x\to0} -x \left[\frac{\sin x}{x}\right]^2\frac 1{\cos x} = 0 \cdot 1^2 \cdot 1 = 0
$$
So
$$\ln y \to 0 \Rightarrow y\to 1$$
A: $\sin x \sim x,$ near $0,$ so you want to find the limit of $x \log x.$ Write it (as you did) as $\log x/(1/x),$ and use L'hopital.
A: Or also
$$\lim_{x\rightarrow 0} x^{\frac{\sin x}{x}x}$$
since $\lim_{x\rightarrow 0} \frac{\sin x}{x}=1$ and ${0}^0=1$
A: Write $x^{\sin{x}} = \exp{\frac{\log{x}}{\frac{1}{\sin{x}}}}$ and apply rule of l'hopital:
$$\lim_{x\rightarrow 0}x^{\sin{x}} = \exp{\left(\lim_{x\rightarrow 0}\frac{\log{x}}{\frac{1}{\sin{x}}}\right)}$$
Now (by 2$\times$ l'hopital)$$\lim_{x\rightarrow 0}\frac{\log{x}}{\frac{1}{\sin{x}}} = \lim_{x\rightarrow 0}\frac{\sin{2x}}{ \cos{x}-x\sin{x}} = 0$$
Thus 
$$\lim_{x\rightarrow 0}x^{\sin{x}} = \exp{\left(\lim_{x\rightarrow 0}\frac{\sin{2x}}{ \cos{x}-x\sin{x}}\right)} = \exp{0} = 1$$
A: Rewrite the limit of the logarithm as
$$\lim_{x\to 0^+}\frac{\sin x}{x}x\ln x.$$
The behaviour of $\frac{\sin x}{x}$ is known. So we need $\lim_{x\to 0^+} x\ln x$, which is probably familiar, and yields to one application of L'Hospital's Rule. 
A: Another way to think about this: 
$$\lim_{x \to 0} \sin x \ln x = \lim_{x \to 0} x \ln  x$$
(because $(\sin x) /x \to 1$ as $x \to 0$).  Writing $x = 1/y$,
we have
$$\lim_{x \to 0} x \ln x = - \lim_{y \to +\infty}(\ln y)/y.$$ 
Now $\ln y$ grows slower than any positive power of $y$, and so 
$\ln y / y \to 0$ as $y \to +\infty$.
(Or, one can make another subsitution: $y = e^z,$ so that
$(\ln y)/y = z/e^z,$ and then use the fact that $e^z$ grows faster
than any power of $z$, so that $z/e^z \to 0$ as $z \to + \infty$.)
