Question

The value of $\underset{x\to 0}{lim}\frac{\sin x+\log \left(1-x\right)}{x^2}$ equals

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

Answer 1

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

$\underset{x\to 0}{lim}\frac{\sin x+\log \left(1-x\right)}{x^2}$

$\frac{0}{0}form$

$=\underset{x\to 0}{lim}\frac{\cos x-\frac{1}{1-x}}{2x}$

(L-hospital)

$=\underset{x\to 0}{lim}\frac{\left(1-x\right)\cos x-1}{2x\left(1-x\right)}$

$=\underset{x\to 0}{lim}\frac{\cos x-x\cos x-1}{2x\left(1-x\right)}$

$\frac{0}{0}form$

$=\underset{x\to 0}{lim}\frac{-\sin x+x\sin x-\cos x}{-2x+2\left(1-x\right)}$

$=\frac{0+0-1}{0+2}$

$=-\frac{1}{2}$