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

$\frac{1}{2}$ 0 1 None of these

$\frac{1}{2}$

$\underset{x→0}{\lim}\frac{x\cos x-\log(1+x)}{x^2}=\underset{x→0}{\lim}\frac{x(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+-...)-(x-\frac{x^2}{2}+\frac{x^3}{3}-+...)}{x^2}$ $\underset{x→0}{\lim}(\frac{1}{2}-\frac{x}{2!}-\frac{x}{3}+....)=\frac{1}{2}$