Question

Evaluate the limit:
$\underset{x\to 0}{lim}\frac{x(2^x-1)}{1-\cos x}$

Answer 1

We have,
$\underset{x\to 0}{lim}\frac{x(2^x-1)}{1-\cos x}$

Dividing the Numerator and Denominator by $x^2$
$\underset{x\to 0}{lim}\frac{\frac{x(2^x-1)}{x^2}}{\frac{1-\cos x}{x^2}}$

$=\underset{x\to 0}{lim}[\left(\frac{2^x-1}{x}\right)\times \frac{x^2}{1-\cos x}]$

$=\underset{x\to 0}{lim}[\left(\frac{2^x-1}{x}\right)\times \frac{x^2}{2{\sin }^2\left(\frac{x}{2}\right)}][\because 1-\cos 2\theta =2{\sin }^2\theta ]$

$=\underset{x\to 0}{lim}[\left(\frac{2^x-1}{x}\right)\times \frac{\frac{x^2}{4}\times 4}{2{\sin }^2\left(\frac{x}{2}\right)}]$

$=\underset{x\to 0}{lim}\left(\frac{2^x-1}{x}\right)\times \underset{x\to 0}{lim}\frac{2}{{\left(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)}^2}$

$=\log 2\times \frac{2}{1^2}[\underset{x\to 0}{lim}\frac{a^x-1}{x}=\log a,\underset{x\to 0}{lim}\frac{\sin x}{x}=1]$

$=2\log 2=\log (2{)}^2=\log 4$
$[\because N\log M=\log M^N]$

Therefore,
$\underset{x\to 0}{lim}\frac{x(2^x-1)}{1-\cos x}=\log 4$