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

Question

$$\lim_{x\to 0}\frac{x\tan x}{\sqrt{1-x^2}-1}$$

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

$$ \begin{align} \lim_{x\to 0}\frac{x\tan x}{\sqrt{1-x^2}-1}\approx\lim_{x\to 0}\frac{x^2}{\sqrt{1-x^2}-1} &=\lim_{x\to 0}-\left(-\frac{x^2}{\sqrt{1-x^2}-1}\right)\\ &=\lim_{x\to 0}-\left(\frac{1-x^2-1}{\sqrt{1-x^2}-1}\right)\end{align}$$

Answer 1

Take the conjugate and then use $\lim_{x\to 0}\frac{\tan x}{x}=1$,

$$\lim_{x\to 0}\frac{x\tan x}{\sqrt{1-x^2}-1}$$ $$=\lim_{x\to 0}\frac{x\tan x (\sqrt{1-x^2}+1)}{-x^2} $$ $$=-\lim_{x\to 0}\frac{\tan x}x (\sqrt{1-x^2}+1)= -2$$

Answer 2

By using $x=\sin 2u$ we obtain $$L{=\lim_{u\to 0}{\sin 2u\cdot \tan(\sin 2u)\over \cos 2u-1}\\=\lim_{u\to 0}{2\sin u\cos u\cdot \tan(\sin 2u)\over -2\sin^2u}\\=-\lim_{u\to 0}{\tan (\sin 2u)\over \sin u}\\=-2}$$since $\tan w\approx w$ and $\sin u\approx u$ for $|u|<<1$.

Answer 3

For small u, $\ \sqrt{1-u} \approx 1-\frac{1}{2}u$.

When $x$ is small, so is $x^2$, thus the following approximation is also valid:

$\ \sqrt{1-x^2} \approx 1-\frac{1}{2}x^2$

Also using $\tan x\approx x$, and forgetting rigour (lol), we get:

$\lim_{x\to 0}\frac{x\tan x}{\sqrt{1-x^2}-1} = \lim_{x\to 0}\frac{x^2}{1-1/2 (x^2)-1} = -2$

Answer 4

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 $$-2-\frac{\pi ^2}{216}\approx -2.046$$