Evaluate an trigonometric limit in $0$ Find the limit 
$$\lim_{x \to 0}\frac{\sin(\sqrt{x})}{x}$$
whithout using and using L'Hospital Rule
We have 
$$\frac{\sin(\sqrt{x})}{x}=\frac{\sin(\sqrt{x})}{\sqrt{x}}\times\frac{1}{\sqrt{x}}
 \to 1 \times (+ \infty)=+\infty$$
Is correct is approach?

We know that $\sin x=x-\dfrac{1}{3}x^{3}+\cdots$ for $|x|<1$, by alternating series grouping, one can see that $\sin x\geq x-\dfrac{1}{3}x^{3}$ for all small $x>0$, so considering small $x>0$, we have
\begin{align*}
\frac{\sin(\sqrt{x})}{x}\geq\frac{\sqrt{x}-3^{-1}x^{3/2}}{x}=\frac{1}{\sqrt{x}}-\frac{1}{3}\sqrt{x},
\end{align*}
taking $x\rightarrow 0^{+}$ and use Squeeze Theorem would yield the result.
