Question

Solve: $\underset{x\to 0}{lim}\frac{x{2}^x-x}{1-\cos x}$

Answer 1

$\underset{x\to 0}{lim}\frac{x{2}^x-x}{1-\cos x}=\frac{0}{0}$ form

Using L'Hospital's Rule, we get

$\underset{x\to 0}{lim}\frac{x{2}^x\log 2+2^x-1}{-(-\sin x)}=\frac{0}{0}$ form

Again using L' Hospital's Rule, we get

$\underset{x\to 0}{lim}\frac{x{2}^x(\log 2{)}^2+2^x\log 2+2^x\log 2}{\cos x}$

$=\frac{0+\log 2+\log 2}{1}$

$=2\log 2$

$\therefore \underset{x\to 0}{lim}\frac{x{2}^x-x}{1-\cos x}=2\log 2$.