Question

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

• A. 1
• B.
• C.
• D.

Answer 1

The correct option is B

We have, $\underset{x\to 0}{lim}\frac{x{.2}^x-x}{1-cosx}$

At x=0,the value of the expression is the form $\frac{0}{0}$

applying L'Hospital Rule

Differentiate numerator and denominator and then apply limit.

$\underset{x\to 0}{lim}\frac{\frac{d}{dx}\left(x{.2}^x\right)-\frac{d}{dx}x}{\frac{d}{dx}(1-cosx)}$

= $\underset{x\to 0}{lim}\frac{x.\frac{d}{dx}{2}^x+2^x\frac{d}{dx}x-1}{0-(-sinx)}$ $\left\{\frac{d}{dx}\right(uv)=u.\frac{dv}{dx}+v.\frac{dy}{dx}\}$

$=\underset{x\to 0}{lim}\frac{x{.2}^x.lo{g}_e^2+2^x-1}{sinx}$

Again At x=0,the value of the expression is the form $\frac{0}{0}$

Differentiate numerator and denominator again and then apply limit.