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?
 A: 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:
$$\begin{align}
\frac{1}{\sqrt{1+x}-1}&=\frac{\sqrt{1+x}+1}{x}
\end{align}$$
A: Notice, use rationalization as follows $$\lim_{x\to 0}\frac{\sqrt{1+x}-\sqrt{1-x^2}}{\sqrt{1+x}-1}$$
$$=\lim_{x\to 0}\frac{1+x-(1-x^2)}{1+x-1}\times \left(\frac{\sqrt{1+x}+1}{\sqrt{1+x}+\sqrt{1-x^2}}\right)$$
$$=\lim_{x\to 0}\frac{x^2+x}{x}\left(\frac{\sqrt{1+x}+1}{\sqrt{1+x}+\sqrt{1-x^2}}\right)$$
$$=\lim_{x\to 0}(x+1)\left(\frac{\sqrt{1+x}+1}{\sqrt{1+x}+\sqrt{1-x^2}}\right)$$
$$=(0+1)\left( \frac{\sqrt{1+0}+1}{\sqrt{1+0}+\sqrt{1-0}}\right)$$
$$=1\left(\frac{2}{2}\right)=\color{red}{1}$$
A: 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
$$\dfrac{\sin y-\sin\dfrac\pi4}{\cos y-\sin\dfrac\pi4}=\cdots=-\cot\dfrac{y+\pi/4}2$$
A: 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...
A: 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?
