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 ?
 A: $$\lim_{x\to 0^+}\frac{\sin(x)}{\sqrt{x\sin(4x)}}=\frac12\lim_{x\to 0^+}\frac{\sin(x)}x\sqrt{\frac{4x}{\sin(4x)}}=\frac12,$$
$$\lim_{x\to 0^-}\frac{\sin(x)}{\sqrt{x\sin(4x)}}=-\frac12\lim_{x\to 0^-}\frac{\sin(x)}x\sqrt{\frac{4x}{\sin(4x)}}=-\frac12,$$
because $\sqrt{x^2}=\pm x$.
A: Look at how the limit behaves when $x>0$ then look at how it behaves when $x<0$.  And keep in mind that the result of the square root of a positive real number, as written in your equation, is always defined as positive.
A: $$\begin{align}&x>0:\;\;\frac{\sin x}{\sqrt{x\sin 4x}}=\sqrt\frac{\sin x}x\cdot\frac1{2\sqrt{\cos x\cos2x}}\xrightarrow[x\to0^+]{}1\cdot\frac12=\frac12\\{}\\&x<0:\;\;-\frac{-\sin x}{\sqrt{(-x)(-\sin 4x)}}=-\sqrt\frac{-\sin x}{-x}\cdot\frac1{2\sqrt{(-\cos x)(-\cos2x)}}\xrightarrow[x\to0^-]{}-1\cdot\frac12=-\frac12\end{align}$$
A: First note that $\displaystyle\lim_{x\to0}\frac{\sin x}x=1$.  That's a well-known limit.Because of it you can replace $\sin x$by $x$, and $\sin 4x$by $4x$in your limit.Then your limit becomes
$$ \displaystyle\lim_{x\to0}\frac x{\sqrt{4x^2}}=\tfrac12\lim_{x\to0}\frac x{|x|} $$
But $\displaystyle\lim_{x\to0}\frac x{|x|}$doesn't exist because the right limit is $1$while the left limit is $-1$.
A: 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
$$
\lim_{x\to0}\left|\frac{\sin x}{\sqrt{x\sin 4x}}\right|=
\lim_{x\to0}\sqrt{\frac{\sin^2 x}{x\sin 4x}}=
\lim_{x\to0}\sqrt{\frac{1}{4}\frac{\sin^2 x}{x^2}\frac{4x}{\sin 4x}}=
\frac{1}{2}
$$
A: Let $f(x)=\sin x/\sqrt{x\sin4x}$.  Note that $f$ is an odd function, i.e., $f(-x)=-f(x)$.  Consequently, $\lim_{x\to0}f(x)$, if it exists, must equal $0$.  But if the limit exists, then
$$\left(\lim_{x\to0}f(x)\right)^2=\lim_{x\to0}(f(x))^2=\lim_{x\to0}{\sin^2x\over x\sin4x}={1\over4}\lim_{x\to0}\left(\left(\sin x\over x \right)^2\left(4x\over\sin4x\right)\right)={1\over4}\not=0$$
Thus the limit does not exist.
