## Calculus 8th Edition

$\lim\limits_{x \to 0^{+}}\sqrt x [1+sin^{2}(2\pi/x)]=0$
$\lim\limits_{x \to 0^{+}}\sqrt x [1+sin^{2}(2\pi/x)]$ 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 $\lim\limits_{x \to 0^{+}}\sqrt x [1+sin^{2}(2\pi/x)]=0$