Solve $ \lim _{x\to 0}\left(\frac{\sin x}{\sqrt{1-\cos x}}\right) $ I have to solve 

$$ \lim _{x\to 0}\left(\frac{\sin x}{\sqrt{1-\cos x}}\right) $$

How many times do I have to apply L'Hopital in this?
 A: Note that, near $0$, $\frac{\sin x}{\sqrt{1-\cos x}}$ has the same sign as $x$. Therefore, either your limit does not exist or it is equal to $0$. On the other hand,\begin{align}\lim_{x\to0}\left(\frac{\sin x}{\sqrt{1-\cos x}}\right)^2&=\lim_{x\to0}\frac{\sin^2x}{1-\cos x}\\&=\lim_{x\to0}1+\cos x\\&=2.\end{align}Therefore$$\lim_{x\to0^\pm}\frac{\sin x}{\sqrt{1-\cos x}}=\pm\sqrt2$$and your limit does not exist.
Notice that I applied L'Hopital's rule $0$ times in order to prove this.
A: write $$\frac{\sin(x)\sqrt{1+\cos(x)}}{\sqrt{1-\cos(x)}\sqrt{1+\cos(x)}}$$
Can you finish?
A: 
L'Hospital's Rule doesn't work here as it leads to an endless loop.  See the development following the hint.


HINT:
Since applying L'Hospital's Rule fails, I wanted to provide a hint toward analyzing the limit of interest.
Note that $1-\cos(x)=2\sin^2(x/2)$. 
Judiciously apply the limit, $\lim_{t\to 0}\frac{\sin(t)}{t}=1$.
You will find that the limit fails to exist since the right-side and left-side limits are not equal.

To see the futility of L'Hospital's Rule, we apply LHR once to find
$$\begin{align}
\lim_{x\to 0}\frac{\sin(x)}{\sqrt{1-\cos(x)}}&=\lim_{x\to 0}\frac{\cos(x)}{\frac12 (1-\cos(x))^{-1/2}\,\sin(x)}\\\\
&=\lim_{x\to 0}\frac{2\cos(x)\sqrt{1-\cos(x)}}{\sin(x)}
\end{align}$$
A second application reveals
$$\begin{align}
\lim_{x\to 0}\frac{2\cos(x)\sqrt{1-\cos(x)}}{\sin(x)}&=\lim_{x\to 0}\left(\frac{-2\sin(x)\sqrt{1-\cos(x)}+2\cos(x)\frac12(1-\cos(x))^{-1/2}\,\sin(x)}{\cos(x)}\right)\\\\
&=\lim_{x\to 0}\left(\frac{\sin(x)}{\sqrt{1-\cos(x)}}-2\tan(x)\sqrt{1-\cos(x)}\right)\tag1
\end{align}$$
If the proposed limit exists, then since $\lim_{x\to 0}\left(-2\tan(x)\sqrt{1-\cos(x)}\right)=0$, we find after two applications of LHR that we are back to where we started.
A: $$
\frac{\sin x}{\sqrt{1-\cos x}}=\frac{\sin(x)}{|\sin(x)|}\sqrt{1+\cos(x)}
$$
Thus,
$$
\lim_{x\to0^+}=\frac{\sin x}{\sqrt{1-\cos x}}=\sqrt2
$$
and
$$
\lim_{x\to0^-}=\frac{\sin x}{\sqrt{1-\cos x}}=-\sqrt2
$$
Therefore, since the limits from different directions are different, the limit does not exist.
A: $$ \lim _{x\to 0^{\pm}}\left(\frac{\sin x}{\sqrt{1-\cos x}}\right)$$
$$= \lim _{x\to 0^{\pm}}\left(\frac{2\sin \frac{x}{2}\cos \frac{x}{2}}{\sqrt{2}|\sin \frac{x}{2}|}\right)$$
$$=\pm\sqrt{2}$$
$(\because \sin x=2\sin \frac{x}{2}\cos \frac{x}{2}, 1-\cos x=2\sin^2 \frac{x}{2})$
please note that Here I applied LHR zero times.
A: $$\lim_{x\rightarrow0^+}\frac{\sin{x}}{\sqrt{1-\cos{x}}}=\frac{1}{\sqrt2}\lim_{x\rightarrow0^+}\frac{\sin{x}}{|\sin\frac{x}{2}|}=\sqrt2\lim_{x\rightarrow0^+}\frac{\frac{\sin{x}}{x}}{\frac{\sin\frac{x}{2}}{\frac{x}{2}}}=\sqrt2;$$
$$\lim_{x\rightarrow0^-}\frac{\sin{x}}{\sqrt{1-\cos{x}}}=\frac{1}{\sqrt2}\lim_{x\rightarrow0^-}\frac{\sin{x}}{|\sin\frac{x}{2}|}=-\sqrt2\lim_{x\rightarrow0^-}\frac{\frac{\sin{x}}{x}}{\frac{\sin\frac{x}{2}}{\frac{x}{2}}}=-\sqrt2.$$
A: This limit does not exist so you do not have to use L'Hospital's rule at all. Use half angle formulas for sin(x) and 1-cos(x). Simplify and you will get$\sqrt{2}$ 
 for the right limit and - $\sqrt{2}$ for the left limit. A graph or a table of values will be helpful to see the jump from one side of zero to the other side.   
