Question

$\underset{x\to 0}{lim}\frac{\log (1+x\sin x)}{\cos x-1}$ is

• A. $-2$
• B. 2
• C. $0$
• D. 1

Answer 1

The correct option is A $-2$

$\underset{x\to 0}{lim}\frac{\log (1+xsinx)}{cosx-1}$
It is of the form $\frac{0}{0}$, so applying L-Hospital's rule

$=\underset{x\to 0}{lim}\frac{1}{1+x\sin x}\frac{\sin x+x\cos x}{-\sin x}$
$=1.\underset{x\to 0}{lim}\frac{\sin x+x\cos x}{-\sin x}$
$=\underset{x\to 0}{lim}\frac{\frac{\sin x}{x}+\cos x}{-\frac{\sin x}{x}}$
$=\frac{1+1}{-1}=-2$
Alternative Method:
$\underset{x\to 0}{lim}\frac{\log (1+x\sin x)}{\cos x-1}$
$=\underset{x\to 0}{lim}\frac{\log (1+x\sin x)}{x\sin x}\underset{x\to 0}{lim}\frac{x\sin x}{\cos x-1}$
$=\underset{x\to 0}{lim}\frac{\sin x}{x}\underset{x\to 0}{lim}\frac{x^2}{\cos x-1}$
$=\underset{x\to 0}{lim}\frac{x^2}{2{\sin }^2\frac{x}{2}}$
$=\underset{x\to 0}{lim}\frac{4\left(\frac{x^2}{4}\right)}{-2{\sin }^2\frac{x}{2}}$
$=-2$