How to find the limit $\lim_{x \to 0} \frac{1-\sqrt{1-x^2}}{x}$?

Question

Not sure where to even start at this one. I think I could get it to this form:

$$\lim_{x \to 0} \frac{1-\sqrt{(x+1)(x-1)}}{x}$$

Why cant I do this any further? EDIT: That square root really messes me up often and I get confused about what to do next when I see it

Answer 1

\begin{eqnarray} \lim_{x\to0}\frac{1-\sqrt{1-x^2}}{x}&=&\lim_{x\to 0}\frac{(1-\sqrt{1-x^2})(1+\sqrt{1-x^2})}{x(1+\sqrt{1-x^2})}=\lim_{x\to 0}\frac{1-(1-x^2)}{x(1+\sqrt{1-x^2})}\\ &=&\lim_{x\to0}\frac{x^2}{x(1+\sqrt{1-x^2})}=\lim_{x\to0}\frac{x}{1+\sqrt{1-x^2}}=0. \end{eqnarray}

Answer 2

$$ \frac{1-\sqrt{1-x^2}}{x}= \frac{x^2}{x(1+\sqrt{1-x^2})} $$ Can you take it from there?