Finding $\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x$ I need to compute a limit:
$$\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x$$
I tried to apply the L'Hôpital rule, but the emerging terms become too complicated and doesn't seem to simplify.
$$
\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x \\
= \exp (\lim_{x \to 0+} x \ln (2\sin \sqrt x + \sqrt x \sin \frac{1}{x})) \\
= \exp (\lim_{x \to 0+} \frac 
{\ln (2\sin \sqrt x + \sqrt x \sin \frac{1}{x})} 
{\frac 1 x}) \\
= \exp \lim_{x \to 0+} \dfrac 
{\dfrac {\cos \sqrt x} {x} + \dfrac {\sin \dfrac 1 x} {2 \sqrt x} 
- \dfrac {\cos \dfrac 1 x} {x^{3/2}}} 
{- \dfrac {1} {x^2} \left(2\sin \sqrt x + \sqrt x \sin \frac{1}{x} \right)} 
$$
I've calculated several values of this function, and it seems to have a limit of $1$.

For small positive angles, the inequalities ${3\over4}\theta\le\sin\theta\le\theta$ apply. (The fraction $3/4$ is somewhat arbitrary; any fraction less than $1$ will do.)  Letting $\theta=\sqrt x$ and noting that $-1\le\sin(1/x)\le1$ for all $x\not=0$, we have
$${3\over2}\sqrt x-\sqrt x\le2\sin\sqrt x+\sqrt x\sin(1/x)\le2\sqrt x+\sqrt x$$
which simplifies to
$${1\over2}\sqrt x\le2\sin\sqrt x+\sqrt x\sin(1/x)\le3\sqrt x$$
If we take for granted the limit $x^x\to1$ as $x\to0$ (easily proved with L'Hopital applied to $\ln x\over1/x$), we have $\lim_{x\to0^+}(\sqrt x/2)^x=\lim_{x\to0^+}(3\sqrt x)^x=1$ so the Squeeze Theorem tells us
$$\lim_{x\to0^+}(\sin\sqrt x+\sqrt x\sin(1/x))^x=1$$
