Evaluate $\lim_{x\to 0}\frac{x\tan x}{\sqrt{1-x^2}-1}$ Without Lhopital $$\lim_{x\to 0}\frac{x\tan x}{\sqrt{1-x^2}-1}$$


Tried using $\tan x\approx x$ and $\lim_{x\to 0}\frac{\tan x}{x}=1$

$$
\begin{align} \lim_{x\to 0}\frac{x\tan x}{\sqrt{1-x^2}-1}\approx\lim_{x\to 0}\frac{x^2}{\sqrt{1-x^2}-1}  &amp;=\lim_{x\to 0}-\left(-\frac{x^2}{\sqrt{1-x^2}-1}\right)\\  &amp;=\lim_{x\to 0}-\left(\frac{1-x^2-1}{\sqrt{1-x^2}-1}\right)\end{align}$$
Assumin that you could like more than the limit itself, consider the composition of Taylor series
$$\sqrt{1-x^2}=1-\frac{x^2}{2}-\frac{x^4}{8}+O\left(x^6\right)$$
$$\sqrt{1-x^2}-1=-\frac{x^2}{2}-\frac{x^4}{8}+O\left(x^6\right)$$
$$x\frac{\tan (x)}{\sqrt{1-x^2}-1}=x\frac{x+\frac{x^3}{3}+\frac{2 x^5}{15}+O\left(x^7\right) } {-\frac{x^2}{2}-\frac{x^4}{8}+O\left(x^6\right) }$$ Now, long division to get
$$\frac{x\tan (x)}{\sqrt{1-x^2}-1}=-2-\frac{x^2}{6}+O\left(x^4\right)$$

For the fun of it, try for $x=\frac \pi 6$ whch is quite far away from $0$. The exact value is 
$$\frac{\pi }{\sqrt{3} \left(\sqrt{36-\pi ^2}-6\right)}\approx -2.042$$ while the series would give
$$-2-\frac{\pi ^2}{216}\approx -2.046$$
First of all, we have $\lim_{x\to 0}\frac{x\tan x}{\sqrt{1-x^2}-1}=\lim_{x\to 0}\frac{x^2}{\sqrt{1-x^2}-1}\cdot \frac{\tan x}x=\lim_{x\to 0}\frac{x^2}{\sqrt{1-x^2}-1}.$ Changing variables ($y=-x^2$,) we have
\begin{equation}
\begin{split}
\lim_{x\to 0}\frac{x^2}{\sqrt{1-x^2}-1}&amp;=\lim_{y\to 0^+}-\frac{y}{\sqrt{1+y}-1}\\
&amp;=\lim_{y\to 0^+}-\frac{y(\sqrt{1+y}+1)}{y}\\
&amp;=-2
\end{split}
\end{equation}