$\underset{x\to 0}{Lt}\frac{1+\sin x-\cos x+\log (1-x)}{x^3}$ is equal to

Use LH rule three times to get the limit value

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

use LH rule 

$\underset{x\to 0}{Lt}\frac{\cos x+\sin x-{\displaystyle \frac{1}{1-x}}}{3x^2}$

Again use LH rule

$\underset{x\to 0}{Lt}\frac{-\sin x+\cos x+{\displaystyle \frac{1}{{\left(1-x\right)}^2}}}{6x}$

Again use LH rule

$\underset{x\to 0}{Lt}\frac{-\cos x-\sin x-{\displaystyle \frac{2}{{\left(1-x\right)}^3}}}{6}$

After substituting the limit value

we get the limit value as $\frac{-1-0-2}{6}=\boxed{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}}$