How do you show that $\displaystyle\lim_{x\to 0}\frac{\sin(x)}{\sqrt{x\sin(4x)}} $does not exist? How can I show that $\displaystyle\lim_{x\to 0}\frac{\sin(x)}{\sqrt{x\sin(4x)}} $does not exist ?

Since $\sqrt{x\sin(4x)}>0$ in a (punctured) neighborhood of $0$, while $\sin x<0$ for $-\pi<x<0$ and $\sin x>0$ for $0<x<\pi$, the limit exists if and only if it is $0$ or, equivalently,
$$
\lim_{x\to0}\left|\frac{\sin x}{\sqrt{x\sin 4x}}\right|=0
$$
However
$$
\lim_{x\to0}\left|\frac{\sin x}{\sqrt{x\sin 4x}}\right|=
\lim_{x\to0}\sqrt{\frac{\sin^2 x}{x\sin 4x}}=
\lim_{x\to0}\sqrt{\frac{1}{4}\frac{\sin^2 x}{x^2}\frac{4x}{\sin 4x}}=
\frac{1}{2}
$$
