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Question

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

A

$-1$

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B

$\frac{1}{2}$

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C

$-\frac{1}{2}$

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D

$1$

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Solution

## 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}{\mathrm{lim}}\frac{\left[1+\mathrm{sin}\left(x\right)-\mathrm{cos}\left(x\right)+\mathrm{ln}\left(1-x\right)\right]}{{x}^{3}}$$=\frac{1+0-1+0}{0}=\frac{0}{0}form$ Step $2$: Apply L. Hospital's rule.$\begin{array}{rcl}\underset{x\to 0}{\mathrm{lim}}\frac{\mathrm{cos}\left(x\right)+\mathrm{sin}\left(x\right)+\frac{\left(-1\right)}{1-x}}{3{x}^{2}}& =& \frac{1+0-1}{0}\\ & =& \frac{0}{0}\end{array}$Since $\frac{0}{0}$ form, apply again L' Hospital's rule,$\underset{x\to 0}{\mathrm{lim}}\frac{-\mathrm{sin}\left(x\right)+\mathrm{cos}\left(x\right)-\frac{1}{{\left(1-x\right)}^{2}}}{6x}=\frac{0}{0}$Since $\frac{0}{0}$ form, apply again L' Hospital's rule,$\begin{array}{rcl}\underset{x\to 0}{\mathrm{lim}}\frac{-\mathrm{cos}\left(x\right)-\mathrm{sin}\left(x\right)-\frac{2}{{\left(1-x\right)}^{3}}}{6}& =& \frac{-1-0-2}{6}\\ & =& -\frac{1}{2}\end{array}$Therefore the value of $\underset{x\to 0}{\mathrm{lim}}\frac{\left[1+\mathrm{sin}\left(x\right)-\mathrm{cos}\left(x\right)+\mathrm{ln}\left(1-x\right)\right]}{{x}^{3}}$ is $-\frac{1}{2}$.Hence, option (C) is the correct option, i.e. $-\frac{1}{2}$.

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