How to solve following limit I've been struggeling a bit with the following limit:
$\lim\limits_{x \to 0} \frac{a- \sqrt{a^2 - x^2}}{x^2}$
The solution is:
If a < 0 then -$\infty$ .
If a > 0 then $\frac{1}{2a}$
But I don't know how to get there.
Thank you.
 A: $$\begin{align*}
\lim_{x\to0}\frac{a-\sqrt{a^2-x^2}}{x^2}
&= \lim_{x\to0}\frac{a-\sqrt{a^2-x^2}}{x^2}\cdot\frac{a+\sqrt{a^2-x^2}}{a+\sqrt{a^2-x^2}}\\
&= \lim_{x\to0}\frac{x^2}{x^2\left(a+\sqrt{a^2-x^2}\right)}\\
&= \lim_{x\to0}\frac{1}{a+\sqrt{a^2-x^2}}\\
&= \frac{1}{a+\sqrt{a^2}}\\
\end{align*}$$
Now the tricky part is how you simplify $\sqrt{a^2}$ base on the sign of $a$.
A: We have that $$\lim\limits_{x \to 0} \frac{a- \sqrt{a^2 - x^2}}{x^2}=\lim\limits_{x \to 0} \frac{a^2- a^2 + x^2}{x^2(a+ \sqrt{a^2 - x^2})}=\frac{1}{a+|a|}$$ from here you can do
A: if $a < 0,$ then the numerator is close to $-2a$ as $x \to 0$ so that the limit does not exist, or limit is $-\infty.$ but if $a > 0,$ then the numerator also goes to zero as $x \to 0$ so you have to do more work to make sense of ${0 \over 0}$
here is one way to do it: for $a > 0$, we use the binomial theorem 
$(BIG+small)^n = BIG^n + n \ BIG^{n-1} small + \cdots \ $ to approximate
$$(a^2 - x^2)^{1/2} = (a^2)^{1/2} + {1 \over 2}(a^2)^{-1/2}(-x^2) + \cdots = a - {x^2 \over 2a}  +\cdots$$
now put this back in the quotient we get ${a - \sqrt{a^2 - x^2} \over x^2 } = {1 \over 2a} + \cdots \to {1 \over 2a}$ as $x \to 0$ because $\cdots$ starts with a multiple of $x^4.$
