Evaluate $\lim_{x \to 0}\frac{x}{\sqrt{1-\sqrt{1-x^2}}}$ Evaluate $$L=\lim_{x \to 0}\frac{x}{\sqrt{1-\sqrt{1-x^2}}} \tag{1}$$
I have used L Hopital's  Rule we get
$$L=\lim_{x \to 0} 2\frac{ \sqrt{1-\sqrt{1-x^2}} \sqrt{1-x^2}}{x}$$
$\implies$
$$L=2 \lim_{x \to 0}\frac{ \sqrt{1-\sqrt{1-x^2}}}{x}$$  and from $(1)$ we get
$$L=\frac{2}{L}$$
$$L=\sqrt{2}$$
is this correct appraoach?
 A: $$\lim_{x\to 0}\frac{x}{\sqrt{1-\sqrt{1-x^2}}}\stackrel{x\mapsto\sin\theta}{=}\lim_{\theta\to 0}\frac{\sin\theta}{\sqrt{1-\cos\theta}}=\lim_{\theta\to 0}\frac{\sin\theta}{\sqrt{2}\left|\sin\frac{\theta}{2}\right|}\stackrel{\theta\mapsto2\varphi}{=}\sqrt{2}\lim_{\varphi\to 0}\frac{\cos\varphi\sin\varphi}{\left|\sin\varphi\right|} $$
clearly does not exist, but $\lim_{x\to 0^{\pm}}(\ldots)=\pm\sqrt{2}$.
A: Your result is wrong, sorry: the limit doesn't exist.
If you want to consider the limit from the right, you can use $x=\sqrt{x^2}$, for $x>0$, and write it as
$$
\lim_{x\to0^+}\sqrt{\frac{x^2}{1-\sqrt{1-x^2}}}=
\lim_{x\to0^+}\sqrt{\frac{x^2(1+\sqrt{1-x^2})}{1-(1-x^2)}}=
\lim_{x\to0^+}\sqrt{1+\sqrt{1-x^2}}=\sqrt{2}
$$
For the limit from the left, you have to take into account that $x=-\sqrt{x^2}$ for $x<0$, so you have
$$
\lim_{x\to0^-}-\sqrt{\frac{x^2}{1-\sqrt{1-x^2}}}=-\sqrt{2}
$$
with the same computations as above.
Your argument that $L=2/L$ just shows that if the limit $L$ exists then $L^2=2$, but tells you nothing about the existence.
A: Your solution is not correct.  If the limit existed, then
$$L=\lim_{x\to0}\frac{-x}{\sqrt{1-\sqrt{1-(-x)^2}}}=-L\\\implies L=0$$
What you have found was
$$\lim_{x\to0^+}\frac x{\sqrt{1-\sqrt{1-x^2}}}=\sqrt2$$
So in fact, the limit doesn't exist.
A: By symmetry, if the limit exists it must be zero (because $x\to 0$ is equivalent to $(-x)\to 0$, so you could replace $x$ by $-x$ and obtain the opposite result, which would have to equal the original result and so be zero). So, in fact, you have shown that the limit does not exist.
A: Alternatively: 
$$L=\lim_\limits{x\to 0} \frac{sgn(x) \sqrt{x^2}}{\sqrt{1-\sqrt{1-x^2}}}=$$
$$\lim_\limits{x\to 0} sgn(x) \sqrt{\lim_\limits{x\to 0} \frac{x^2}{\sqrt{1-\sqrt{1-x^2}}}} =(LR)= $$
$$\lim_\limits{x\to 0} sgn(x) \sqrt{\lim_\limits{x\to 0} 2\sqrt{1-x^2}}=$$
$$\lim_\limits{x\to 0} sgn(x) \sqrt{2}.$$
Hence, the limit exists if $x$ approaches $0$ from left or right, otherwise it does not exist.
A: There is a right limit and a left limit and they are different, which means that the limit doesn't exist because when it exists it is unique.
Let substitute $1-x^2=w^2$
Suppose $x>0$ we have $x=\sqrt{1-w^2}$ and the limit becomes
$$\lim_{w\to 1} \, \dfrac{\sqrt{1-w^2}}{\sqrt{1-w}}=\lim_{w\to 1} \, \frac{\sqrt{1-w} \sqrt{w+1}}{\sqrt{1-w}}=\lim_{w\to 1} \,\sqrt{w+1}= \sqrt{2}$$
If $x<0$ then we have $x=-\sqrt{1-w^2}$ 
And the previous limit gives $- \sqrt{2}$
Hope this helps
