Question

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

• A. $2\log 2$
• B. $\log 2$
• C. $\frac{1}{2}\log 2$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $2\log 2$

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

Using L'Hospital's rule, we differentiate the numerator and denominator to get

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

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

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

$=2\log 2$