Question

The value of $\displaystyle\lim _{x \rightarrow 0} \frac{\left(e^x-x-1\right)(x-\sin x) \ln (1+x)}{x^6}$ is equal to

• A. $\frac{1}{2}$
• B. $\frac{1}{6}$
• C. $\frac{1}{12}$
• D. $\frac{1}{3}$

Answer 1

Answer: (C)

Let, $L=\displaystyle\lim _{x \rightarrow 0}\left(\frac{e^x-x-1}{x^2}\right)\left(\frac{x-\sin x}{x^3}\right)\left(\frac{\ln (1+x)}{x}\right)$
Applying expansion series, we get,
$L=\displaystyle\lim _{x \rightarrow 0} \frac{\left\{\left(1+x+\frac{x^2}{2 !} \cdots\right)-(x+1)\right\}\left\{x-\left(x-\frac{x^3}{3 !}+\right)\right\}\left\{x-\frac{x^2}{2}+\ldots\right\}}{x^6}$
$=\frac{1}{2} \cdot \frac{1}{6} \cdot 1=\frac{1}{12}$