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?

Without too many rationalizations: the term $\sqrt{1+x}$ appears several times. So, set $\sqrt{1+x}=1+t$, that makes $x=2t+t^2$ and $1-x=1-2t-t^2$; then your limit becomes
$$
\lim_{t\to0}\frac{1+t-(1+t)\sqrt{1-2t-t^2}}{t}=
\lim_{t\to0}(1+t)\cdot\lim_{t\to0}\frac{1-\sqrt{1-2t-t^2}}{t}
$$
provided the limit of the second factor exists; but
$$
\lim_{t\to0}\frac{1-\sqrt{1-2t-t^2}}{t}=
\lim_{t\to0}\frac{1-(1-2t-t^2)}{t(1+\sqrt{1-2t-t^2})}
$$
with the usual rationalization. Can you go on?
