Question

The value of $\underset{x\to 0}{\lim }\frac{\left[1+\sin \left(x\right)-\cos \left(x\right)+\ln (1-x)\right]}{x^3}$ is

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

Answer 1

The correct option is C

$-\frac{1}{2}$

Explanation of the correct option.

Step $1$: Put the values of limit.

Given : $\underset{x\to 0}{\lim }\frac{\left[1+\sin \left(x\right)-\cos \left(x\right)+\ln (1-x)\right]}{x^3}$$=\frac{1+0-1+0}{0}=\frac{0}{0}form$

Step $2$: Apply L. Hospital's rule.

$\begin{array}{rcl}\underset{x\to 0}{\lim }\frac{\cos \left(x\right)+\sin \left(x\right)+{\displaystyle \frac{(-1)}{1-x}}}{3x^2}& =& \frac{1+0-1}{0}\\ & =& \frac{0}{0}\end{array}$

Since $\frac{0}{0}$ form, apply again L' Hospital's rule,

$\underset{x\to 0}{\lim }\frac{-\sin \left(x\right)+\cos \left(x\right)-{\displaystyle \frac{1}{{\left(1-x\right)}^2}}}{6x}=\frac{0}{0}$

Since $\frac{0}{0}$ form, apply again L' Hospital's rule,

$\begin{array}{rcl}\underset{x\to 0}{\lim }\frac{-\cos \left(x\right)-\sin \left(x\right)-{\displaystyle \frac{2}{{\left(1-x\right)}^3}}}{6}& =& \frac{-1-0-2}{6}\\ & =& -\frac{1}{2}\end{array}$

Therefore the value of $\underset{x\to 0}{\lim }\frac{\left[1+\sin \left(x\right)-\cos \left(x\right)+\ln (1-x)\right]}{x^3}$ is $-\frac{1}{2}$.

Hence, option (C) is the correct option, i.e. $-\frac{1}{2}$.