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?

Your argumentation is completely right until in the last step you claim that the second term is divergent which is not true. You rather have:
$$\frac{1-\sqrt{1-x^2}}{\sqrt{1+x}-1}=\frac{x^2(\sqrt{1+x}+1)}{x(1+\sqrt{1-x^2}}=x\frac{\sqrt{1+x}+1}{1+\sqrt{1-x^2}}$$
Now, the second factor clearly goes to 1 as $x$ approaches 0 and hence the whole term converges to 0 which makes your original term converge to $1+0=1$.
Note that your idea of rationalizing numerator and denominator was indeed a good approach...
