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 $L=4$
 A: 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$$\begin{align}&\lim_{x\to 0}\frac{\sqrt{2(2-x)}\ (1-\sqrt{1-x^2})}{\sqrt{1-x}\ (2-\sqrt{4-x^2})}\\&=\lim_{x\to 0}\frac{\sqrt{2(2-x)}\ \color{red}{(1-\sqrt{1-x^2})}}{\sqrt{1-x}\ \color{blue}{(2-\sqrt{4-x^2})}}\cdot\frac{\color{red}{1+\sqrt{1-x^2}}}{1+\sqrt{1-x^2}}\cdot\frac{2+\sqrt{4-x^2}}{\color{blue}{2+\sqrt{4-x^2}}}\\&=\lim_{x\to 0}\frac{\sqrt{2(2-x)}\ (2+\sqrt{4-x^2})\color{red}{(1-(1-x^2))}}{\sqrt{1-x}\ (1+\sqrt{1-x^2})\color{blue}{(4-(4-x^2))}}\\&=\lim_{x\to 0}\frac{\sqrt{2(2-x)}\ (2+\sqrt{4-x^2})}{\sqrt{1-x}\ (1+\sqrt{1-x^2})}\\&=\frac{\sqrt 4\ (2+\sqrt 4)}{1\cdot (1+1)}\\&=4\end{align}$$
A: 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 :
$$\lim_{x \to 0} \frac{\sqrt{2(2-x)} (1-\sqrt{1-x^2})}{\sqrt{1-x}(2-\sqrt{4-x^2})}=\\\lim_{x \to 0} \frac{\sqrt{2(2-x)} (1-(1-\frac{1}{2}x^2))}{\sqrt{1-x}(2-(2-\frac{x^2}{4}))} =\\ \lim_{x \to 0} \frac{\sqrt{2(2-x)} (\frac{1}{2}x^2)}{\sqrt{1-x}(\frac{x^2}{4})}=\\\lim_{x \to 0} \frac{\sqrt{2(2-x)} 2}{\sqrt{1-x}}=4 $$
A: $\lim\limits_{x\to 0}\left(\frac{\sqrt{2(2-x)}(1-\sqrt{1-x^2})}{\sqrt{1-x}(2-\sqrt{4-x^2})}\right) = \lim\limits_{x\to 0}\left(\frac{\sqrt{2(2-x)}(2 + \sqrt{4-x^2})}{\sqrt{1-x}(1+\sqrt{1-x^2})}\right) = 4$
A: 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 $a-b = \frac {a^2 - b^2} {a+b}$.)
A: 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 $\sin y\to0$
A: L'Hopital can be used, but it goes a lot better if you notice one thing: There are garbage terms in there that can be factored out, leaving a much cleaner problem. Here's what I'm talking about:
$$\lim_{x\to 0} \frac{\sqrt {2(2-x)}}{\sqrt {1-x}} = \frac{2}{1}=2.$$
So we can move that junk out of way, and mulitply by $2$ at the end. We're left thinking about
$$\lim_{x\to 0}\frac{1-\sqrt {1-x^2}}{2- \sqrt {4-x^2}},$$
which Monsieur L'Hopital can handle well.
