The value of $\underset{x\to 0}{lim} \frac{1+\sin x-\cos x+\log (1-x)}{x^3}$ is

Solution:

We have,

$\begin{array}{l}\underset{x\to 0}{lim} \frac{1+\sin x-\cos x+\log (1-x)}{x^3}\\ \\ \left.=\underset{x\to 0}{lim} \frac{1+\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots \right)-\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}\dots \right)+\left(-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}\right)}{x^3}\right)\\ =\underset{x\to 0}{lim} \left(-\frac{1}{3!}-\frac{1}{3}\right)+x^2\left(\frac{1}{5!}-\frac{1}{5}\right)+\dots =-\frac{1}{6}-\frac{1}{3}=-\frac{1}{2}\end{array}$