Does this limit $\underset{x\to \pi }{\text{lim}}\frac{\sqrt{1-\cos ^2(x)}}{\sin (x)} $ exist? I want to compute this limit
$$        \underset{x\to \pi }{\text{lim}}\frac{\sqrt{1-\cos ^2(x)}}{\sin (x)}   $$
which is one of the indeterminate forms, $\frac00$; Using L'hopital, I get
$$        \underset{x\to \pi }{\text{lim}}\frac{\cos(x)\sin(x)}{\cos(x) \sqrt{1-\cos ^2(x) }}= \underset{x\to \pi }{\text{lim}}\frac{\sin(x)}{ \sqrt{1-\cos ^2(x)}}=\frac00    $$
which is again $\frac00$; then, does this mean that the limit does not exist? or there are other ways to compute such a limit that I am not aware of. I appreciate any comments.
 A: When l'Hopital's rule fails, you cannot make any conclusions. You must try something else.
Of course, why apply l'Hopital's rule at all, when you can use identities to simplify the problem?
With the edited version of your question, let's simplify first:
$$\frac{\sqrt{1-\cos^2(x)}}{\sin(x)} = \frac{\sqrt{\sin^2(x)}}{\sin(x)} = \frac{|\sin(x)|}{\sin(x)} = \begin{cases}
+1 &\quad\text{if $0 < x < \pi$} \\
-1 &\quad\text{if $\pi < x < 2\pi$}
\end{cases}
$$
Therefore, the limit as $x$ approaches $\pi$ from below equals $+1$ and the limit as $x$ approaches $\pi$ form above equals $-1$. So, the limit does not exist.
A: First note that
$$
\frac{\sqrt{1-\cos^2 x}}{\sin x} = \frac{\sqrt{\sin^2 x}}{\sin x} = \frac{\left|\sin x \right|}{\sin x }$$
whenever $\sin x \ne 0$.
We check the one-sided limits:
$$\lim_{x \to \pi^+} \frac{\left|\sin x \right|}{\sin x } = -1,$$
$$\lim_{x \to \pi^-} \frac{\left|\sin x \right|}{\sin x } = 1.$$
Therefore the limit $\lim_{x \to \pi} \frac{\sqrt{1-\cos^2 x }}{\sin x }$ does not exist!
