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 $1$.

$$ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ... $$
$$\therefore \ (2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x$$
$$ = \left(2 \left(x^{1/2} - \frac{x^{3/2}}{3!} + \frac{x^{5/2}}{5!} -...\right) + x^{1/2} \sin \frac{1}{x}\right)^x$$
$$ = \left(x^{1/2} \left(2 + \sin \frac{1}{x} - \frac{2x^{3/2}}{3!} + \frac{2x^{5/2}}{5!} -...\right) \right)^x$$
$$ = \left(x^{1/2}\right)^x \left(2 + \sin \frac{1}{x} - \frac{2x}{3!} + \frac{2x^2}{5!} -...\right)^x.$$
Now $ \left(x^{1/2}\right)^x = \left(x^{x}\right)^{1/2} = (e^{x \ln x})^{1/2} = e^{\frac{1}{2} x \ln x},\ $ and $\ \lim\limits_{x\rightarrow 0^{+}}e^{\frac{1}{2}x\ln(x)}=e^{\lim\limits_{x\rightarrow 0^{+}}\frac{1}{2}x\ln(x)}=1. $
Lastly, for $ \left(2 + \sin \frac{1}{x} - \frac{2x}{3!} + \frac{2x^2}{5!} -...\right)^x,\ $  make the substitution $u = \frac{1}{x}$ to get:
$\lim\limits_{u\rightarrow \infty} \left( 2 + \sin u - \frac{2}{3!u} + \frac{2}{5!u^2} - ...\right)^{\frac{1}{u}}$, and the bracket oscillates between $1$ and $3$ for large $u$, so this limit is $1$. Thus:
$$ \lim\limits_{x\rightarrow 0^{+}} \left(x^{1/2}\right)^x \left(2 + \sin \frac{1}{x} - \frac{2x}{3!} + \frac{2x^2}{5!} -...\right)^x = 1 \times 1 = 1.$$
