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 $L=4$

using Bernoulli $$x \to 0 \\ {\color{Red}{(1+ax)^n \approx 1+anx} } \\\sqrt{1-x^2} = (1-x^2)^{\frac{1}{2}} \approx 1-\frac{1}{2}x^2 \\ \sqrt{4-x^2}=\sqrt{4(1-\frac{x^2}{4}})=2(1-\frac{x^2}{4})^{\frac{1}{2}} \approx 2(1-\frac{1}{2} \frac{x^2}{4})=2-\frac{x^2}{4} $$ so by putting them in limit :
$$\lim_{x \to 0} \frac{\sqrt{2(2-x)} (1-\sqrt{1-x^2})}{\sqrt{1-x}(2-\sqrt{4-x^2})}=\\\lim_{x \to 0} \frac{\sqrt{2(2-x)} (1-(1-\frac{1}{2}x^2))}{\sqrt{1-x}(2-(2-\frac{x^2}{4}))} =\\ \lim_{x \to 0} \frac{\sqrt{2(2-x)} (\frac{1}{2}x^2)}{\sqrt{1-x}(\frac{x^2}{4})}=\\\lim_{x \to 0} \frac{\sqrt{2(2-x)} 2}{\sqrt{1-x}}=4 $$
