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?

Set $x=\cos2y$
$$\lim_{x \to 0}\dfrac{\sqrt{1+x}-\sqrt{1-x^2}}{\sqrt{1+x}-1} =\lim_{y\to\pi/4}\dfrac{\sqrt2\cos y-\sin2y}{\sqrt2\cos y-1}$$
$$=-\sqrt2\lim_{y\to\pi/4}\cos y\cdot\lim_{y\to\pi/4}\dfrac{\sin y-\sin\dfrac\pi4}{\cos y-\sin\dfrac\pi4}$$
Method$\#1:$
$$\lim_{y\to\pi/4}\dfrac{\sin y-\sin\dfrac\pi4}{\cos y-\sin\dfrac\pi4} =\dfrac{\lim_{y\to\pi/4}\dfrac{\sin y-\sin\dfrac\pi4}{y-\dfrac\pi4}}{\lim_{y\to\pi/4}\dfrac{\cos y-\cos\dfrac\pi4}{y-\dfrac\pi4}} =\dfrac{\dfrac{d(\sin y)}{dy}_{(\text{ at } y=\pi/4)}}{\dfrac{d(\cos y)}{dy}_{(\text{ at } y=\pi/4)}}$$
Method$\#2:$  Use Prosthaphaeresis Formulas to get
$$\dfrac{\sin y-\sin\dfrac\pi4}{\cos y-\sin\dfrac\pi4}=\cdots=-\cot\dfrac{y+\pi/4}2$$
