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 $x\in\left(0,\frac\pi2\right]$, the concavity of $\sin(x)$ says
$$
\frac2\pi\le\frac{\sin(x)}x\le1
$$
Therefore,
$$
\underbrace{\left(\frac4\pi\sqrt{x}-\sqrt{x}\right)^x}_{\left(\frac4\pi-1\right)^x\sqrt{x^x}}\le\left(2\sin\left(\sqrt{x}\right)+\sqrt{x}\sin\left(\frac1x\right)\right)^x\le\underbrace{\left(2\sqrt{x}+\sqrt{x}\right)^x}_{3^x\sqrt{x^x}}
$$
The Squeeze Theorem says
$$
\lim_{x\to0^+}\left(2\sin\left(\sqrt{x}\right)+\sqrt{x}\sin\left(\frac1x\right)\right)^x=1
$$
