Question

Let $\underset{x\to 0}{\lim }\frac{x\sin x+\log (1-x{)}^x}{x^3}=\frac{m}{n}$ where G.C.D. $(m,n)=1$ then find $|mn|$.

Answer 1

Answer: 2

Let $L=\underset{x\to 0}{\lim }\frac{x\sin x+\log (1-x{)}^x}{x^3}$
$L=\underset{x\to 0}{\lim }\left(\frac{x\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots \right)+x\left(-x-\frac{x^2}{2}-\frac{x^3}{3}-\dots \right)}{x^3}\right)$
$=\underset{x\to 0}{\lim }(-\frac{1}{2}+$ terms containing $x$ and powers of $x)$
$\Rightarrow L=-\frac{1}{2}$
$\Rightarrow m=-1,n=2$
or $m=1,n=-2$
$\Rightarrow |mn|=2$