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 $1$.

Hint: Using the fact that $\frac {\sin x} x \to 1$ as $x \to 0$ verify that $ x\ln (\frac 1  2 \sqrt  x) <x\ln (2\sin\sqrt x+\sqrt x \sin (\frac 1 x)) <x \ln (3\sqrt x)$. Conclude that the limit is $e^{0}=1$.
[$x \ln x \to 0$ as $x \to 0+$]
