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 $1$.
 A: 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 $x \to 0+$]
A: $$ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ... $$
$$\therefore \ (2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x$$
$$ = \left(2 \left(x^{1/2} - \frac{x^{3/2}}{3!} + \frac{x^{5/2}}{5!} -...\right) + x^{1/2} \sin \frac{1}{x}\right)^x$$
$$ = \left(x^{1/2} \left(2 + \sin \frac{1}{x} - \frac{2x^{3/2}}{3!} + \frac{2x^{5/2}}{5!} -...\right) \right)^x$$
$$ = \left(x^{1/2}\right)^x \left(2 + \sin \frac{1}{x} - \frac{2x}{3!} + \frac{2x^2}{5!} -...\right)^x.$$
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:
$$ \lim\limits_{x\rightarrow 0^{+}} \left(x^{1/2}\right)^x \left(2 + \sin \frac{1}{x} - \frac{2x}{3!} + \frac{2x^2}{5!} -...\right)^x = 1 \times 1 = 1.$$
A: Hint.
For $x>0$ small we have
$$
\left(2\sin(\sqrt{x})-\sqrt{x}\right)^x\le \sigma(x)\le \left(2\sin(\sqrt{x})+\sqrt{x}\right)^x
$$

A: 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
$$\lim_{x\to0^+}(\sin\sqrt x+\sqrt x\sin(1/x))^x=1$$
A: 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
$$
\lim_{x\to0^+}\left(2\sin\left(\sqrt{x}\right)+\sqrt{x}\sin\left(\frac1x\right)\right)^x=1
$$
