Answer
$\lim\limits_{x \to 0^{+}}\sqrt x [1+sin^{2}(2\pi/x)]=0$
Work Step by Step
$\lim\limits_{x \to 0^{+}}\sqrt x [1+sin^{2}(2\pi/x)]$
Lets estimate the function:
$-1\leq sin (2\pi/x) \leq 1$
therefore
$-1\leq sin^{2}(2\pi/x) \leq 1$
therefore
$0\leq 1+ sin^{2}(2\pi/x) \leq 2$
$0\leq \sqrt x[1+ sin^{2}(2\pi/x)] \leq 2\sqrt x$
$\lim\limits_{x \to 0^{+}}2\sqrt x=2\times0=0$
$\lim\limits_{x \to 0^{+}}0=0$
By the Squeeze Theorem
$0\leq \sqrt x[1+ sin^{2}(2\pi/x)] \leq 2\sqrt x$
and
$\lim\limits_{x \to 0^{+}}2\sqrt x=\lim\limits_{x \to 0^{+}}0=0$
then
$\lim\limits_{x \to 0^{+}}\sqrt x [1+sin^{2}(2\pi/x)]=0$
