<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] 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
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