Question

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

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

Answer 1

The correct option is D $\frac{1}{2}$

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