Question

Evaluate the limit:
$\underset{x\to 0}{lim}\frac{e^x-1}{\sqrt{1-\cos x}}$

Answer 1

We have,
$\underset{x\to 0}{lim}\frac{e^x-1}{\sqrt{1-\cos x}}[\because 1-\cos 2\theta =2{\sin }^2\theta ]$

$=\underset{x\to 0}{lim}\frac{e^x-1}{\sqrt{2{\sin }^2\frac{x}{2}}}$

$=\underset{x\to 0}{lim}\frac{e^x-1}{\sqrt{2}\left|\sin \frac{x}{2}\right|}$

Dividing the Numerator and Denominator by 𝑥
$=\underset{x\to 0}{lim}\frac{e^x-1}{x}\times \frac{1}{\frac{\sqrt{2}\left|\sin \frac{x}{2}\right|}{2\times \frac{x}{2}}}$

$=\underset{x\to 0}{lim}\frac{e^x-1}{x}\times \frac{\sqrt{2}}{\frac{\left|\sin \frac{x}{2}\right|}{\frac{x}{2}}}$
$=\log e\times \underset{x\to 0}{lim}\frac{\sqrt{2}}{\frac{\left|\sin \frac{x}{2}\right|}{\frac{x}{2}}}\cdots \left(1\right)[\because \underset{x\to 0}{lim}\left(\frac{a^x-1}{x}\right)=\log a]$

Now,
L.H.L. at $x=0$
Put $x=0-h$, in equation $\left(1\right)$
If $x\to 0$, then $h\to 0$
$=\underset{h\to 0}{lim}\frac{\sqrt{2}}{\frac{\left|\sin \frac{-h}{2}\right|}{\frac{-h}{2}}}$

$=\underset{h\to 0}{lim}\frac{-\sqrt{2}}{\frac{\sin \frac{h}{2}}{\frac{h}{2}}}$
$=-\sqrt{2}$

Now,
R.H.L. at $x=0$
Put $x=0+h$, in equation $\left(1\right)$
If $x\to 0$, then $h\to 0$
$=\underset{h\to 0}{lim}\frac{\sqrt{2}}{\frac{\left|\sin \frac{h}{2}\right|}{\frac{h}{2}}}$

$=\underset{h\to 0}{lim}\frac{\sqrt{2}}{\frac{\sin \frac{h}{2}}{\frac{h}{2}}}$
$=\sqrt{2}$

$\text{L.H.L.}\ne \text{R.H.L.}$
Hence, limit does not exist.