<s>[INST] Find $\lim\limits_{x\to 0}\frac{\sqrt{2(2-x)}(1-\sqrt{1-x^2})}{\sqrt{1-x}(2-\sqrt{4-x^2})}$ I use L'Hospitals rule, but can't get the correct limit.
Derivative of numerator in function is
$$\frac{-3x^2+4x-\sqrt{1-x^2}+1}{\sqrt{(4-2x)(1-x^2)}}$$
and derivative of denominator is
$$\frac{-3x^2+2x-2\sqrt{4-x^2}+4}{2\sqrt{(1-x)(4-x^2)}}$$
Now, L'Hospitals rule must be applied again. Is there some easier way to compute the limit?
Limit should be $L=4$
 [/INST] Multiplying it by $$\frac{{1+\sqrt{1-x^2}}}{1+\sqrt{1-x^2}}\cdot\frac{2+\sqrt{4-x^2}}{{2+\sqrt{4-x^2}}}\ (=1)$$gives$$\begin{align}&\lim_{x\to 0}\frac{\sqrt{2(2-x)}\ (1-\sqrt{1-x^2})}{\sqrt{1-x}\ (2-\sqrt{4-x^2})}\\&=\lim_{x\to 0}\frac{\sqrt{2(2-x)}\ \color{red}{(1-\sqrt{1-x^2})}}{\sqrt{1-x}\ \color{blue}{(2-\sqrt{4-x^2})}}\cdot\frac{\color{red}{1+\sqrt{1-x^2}}}{1+\sqrt{1-x^2}}\cdot\frac{2+\sqrt{4-x^2}}{\color{blue}{2+\sqrt{4-x^2}}}\\&=\lim_{x\to 0}\frac{\sqrt{2(2-x)}\ (2+\sqrt{4-x^2})\color{red}{(1-(1-x^2))}}{\sqrt{1-x}\ (1+\sqrt{1-x^2})\color{blue}{(4-(4-x^2))}}\\&=\lim_{x\to 0}\frac{\sqrt{2(2-x)}\ (2+\sqrt{4-x^2})}{\sqrt{1-x}\ (1+\sqrt{1-x^2})}\\&=\frac{\sqrt 4\ (2+\sqrt 4)}{1\cdot (1+1)}\\&=4\end{align}$$
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