Question

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

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. $0$
• D. $2$

Answer 1

The correct option is B $-\frac{1}{2}$

$\underset{x\to 0}{lim}\frac{1+\sin x-\cos x+{\log }_e(1-x)}{x^3}$
Using expansion:
$=\underset{x\to 0}{lim}\frac{1+(x-\frac{x^3}{3!}+\cdots )-(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots )+(-x-\frac{x^2}{2}-\frac{x^3}{3}-\cdots )}{x^3}$

Collecting the coefficients of $x^3$
$=-\frac{1}{2}$