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 $L=4$

For $a>0,$
$$\lim_{x\to0}\dfrac{a-\sqrt{a^2-x^2}}{x^2}=\lim_{x\to0}\dfrac{a^2-(a^2-x^2)}{x^2(a+\sqrt{a^2-x^2})}=\dfrac1{2a}$$ as $x\ne0$ as $x\to0$
Or set $x=a\sin2y\implies y\to0\implies\sin y\to0$
$$\lim_{x\to0}\dfrac{a-\sqrt{a^2-x^2}}{x^2}=a\lim_{y\to0}\dfrac{1-\cos2y}{(a\sin2y)^2}=\dfrac1a\lim_{y\to0}\dfrac{1-\cos2y}{(2\sin y\cos y)^2}$$
$$=\dfrac1a\lim_{y\to0}\dfrac{2\sin^2y}{(2\sin y\cos y)^2}=\dfrac1{2a}$$ as $\sin y\ne0$ as $\sin y\to0$
