Question

$\underset{x\to 0}{lim}\frac{x\cos x-\log (1+x)}{x^2}$ equals

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

Answer 1

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

Substituting gives us indeterminate form.
So, applying L'Hospitals Rule,
${lim}_{x\to 0}\frac{-x\sin x+\cos x-\frac{1}{1+x}}{2x}$
Substituting gives us indeterminate form so, applying L'Hospitals Rule again,
$={lim}_{x\to 0}\frac{-x\cos x-\sin x-\sin x+\frac{1}{1+x}}{2}$
$=\frac{0-0-0+\frac{1}{1+0}}{2}$
$=\frac{1}{2}$
Hence, option 'A' is correct.