How do I solve $\lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x^2}}{\sqrt{1+x}-1}$ indeterminate limit without the L'hospital rule? I've been trying to solve this limit without L'Hospital's rule because I don't know how to use derivates yet. So I tried rationalizing the denominator and numerator but it didn't work.
$\lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x^2}}{\sqrt{1+x}-1}$
What is wrong with $\lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x^2}-1+1}{\sqrt{1+x}-1} = \lim_{x \to 0} \frac{\sqrt{1+x}-1}{\sqrt{1+x}-1} + \lim_{x \to 0} \frac{1-\sqrt{1-x^2}}{\sqrt{1+x}-1}$ = 1 + DIV?

HINT 1:
$$\begin{align}
\sqrt{1+x}-\sqrt{1-x^2}&=\sqrt{1+x}\left(1-\sqrt{1-x}\right)\\\\
&=\sqrt{1+x}\,\,\left(\frac{x}{1+\sqrt{1-x}}\right)
\end{align}$$
HINT 2:
$$\begin{align}
\frac{1}{\sqrt{1+x}-1}&=\frac{\sqrt{1+x}+1}{x}
\end{align}$$
