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 $L=4$

First note that the factors $\sqrt {2 (2-x)}$ and $\sqrt {1-x}$ tend to $2$ and $1$ respectively, therefore they don't raise any problem. It remains to compute the limit of $\frac {1 - \sqrt {1-x^2}} {2 - \sqrt {4-x^2}}$. Note that this fraction can be rewritten as $\frac {1 - (1-x^2)} {1 + \sqrt {1-x^2}} \frac {2 + \sqrt {4 - x^2}} {4 - (4-x^2)} = \frac {2 + \sqrt {4 - x^2}} {1 + \sqrt {1-x^2}}$ which clearly tends to $2$. Putting all the pieces together, the limit is $4$.
(The core fact was that $a-b = \frac {a^2 - b^2} {a+b}$.)
