Question

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

Answer 1

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