Question

Evaluate : $\underset{x\to 0}{\lim }\left(\frac{x.2^x-x}{1-\cos x}\right)$

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

Answer 1

The correct option is A

$2\log 2$

Explanation for the correct option:

Determine the value of$\underset{x\to 0}{\lim }\left(\frac{x.2^x-x}{1-\cos x}\right)$

Consider the given Equation as,

$$\begin{gathered} I=\underset{x\to 0}{\lim }\left(\frac{x.2^x-x}{1-\cos x}\right) \\ \Rightarrow I=\left(\frac{0.2^0-0}{1-\cos 0}\right) \\ \Rightarrow I=\frac{0}{0}\left[\text{indeterminent Form}\right] \end{gathered}$$

Then,

$$\begin{gathered} I=\underset{x\to 0}{\lim }\left(\frac{x.2^x-x}{1-\cos x}\right) \\ \Rightarrow I=\underset{x\to 0}{\lim }\left(\frac{x(2^x-1)}{1-\cos x}\right) \\ \Rightarrow I=\underset{x\to 0}{\lim }\left(\frac{(2^x-1)}{x}\right)\left(\frac{x^2}{1-\cos x}\right)\mathbf{}\mathbf{[}\mathbf{\because }\underset{\mathbf{x}\mathbf{\to }\mathbf{o}}{\mathbf{lim}}\frac{{\mathbf{a}}^{\mathbf{x}}\mathbf{-}\mathbf{1}}{\mathbf{x}}\mathbf{=}\mathbf{lna}\mathbf{,}\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{\mathbf{1}\mathbf{-}\mathbf{cosx}}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{]} \\ \Rightarrow I=\ln \left(2\right)\times 2 \\ \Rightarrow I=2\ln \left(2\right) \end{gathered}$$

Hence, the correct answer is option A.