<s>[INST] 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?
 [/INST] 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.
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