Evaluating the right limit of: $1+\sin(\frac{2\pi}{x}\sqrt{x})$ at $0$ I have a limit as $$\lim_{x\rightarrow 0^{+}} \left((1+\sin\left(\frac{2\pi}{x}\right))\sqrt{x}\right).$$ I am planning to use Squeeze Theorem, so I say that
$-1 \leq \sin(x) \leq 1 \implies -1 \leq \sin(\frac{2\pi}{x}) \leq 1 \implies 0 \leq 1 + \sin\left(\frac{2\pi}{x}\right) \leq 2$ 
$\implies 0 \leq \left(1 + \sin\left(\frac{2\pi}{x}\right)\right)\sqrt{x} \leq 2\sqrt{x}$ 
I use the theorem so I get $\lim_{x\rightarrow 0^{+}} ((1+sin(\frac{2\pi}{x})\sqrt{x})=0$. 
Is there any problem? Also how can I find the limit $\lim_{x \rightarrow 0^{+}}\sin\left(\cfrac{2\pi}{x}\right)$ so then, I evaluate the whole limit without using Squeeze Theorem.


Is there any problem?

Your derivation is fine. Using the squeeze theorem is a good idea since
$$
\lim_{x \rightarrow 0^{+}}\sin\left(\cfrac{2\pi}{x}\right)
$$does not exist. Observe that
$$
\lim_{n\rightarrow \infty}\sin\left(\cfrac{2\pi}{1/n}\right)=\lim_{n\rightarrow \infty}\sin(2n\pi)=0
$$ and that
$$
\lim_{n\rightarrow \infty}\sin\left(\cfrac{2\pi}{1/(n+\frac14)}\right)=\lim_{n\rightarrow \infty}\sin \frac{\pi}2=1\ne0
$$ proving 
$$
\lim_{x \rightarrow 0^{+}}\sin\left(\cfrac{2\pi}{x}\right)
$$
can't exist.
