Question

${\lim }_{x\to 0}\frac{xcosx-log(1+x)}{x^2}$ is equal to

• A. 1/2
• B. 0
• C. 1
• D. none of these

Answer 1

Answer: (A)

${\lim }_{x\to 0}\frac{x\cos x-\log \left(1+x\right)}{x^2}\left(\frac{0}{0}from\right)$
= ${\lim }_{x\to 0}\frac{\cos x-x\sin x-\frac{1}{1+x}}{2x}\left(\frac{0}{0}from\right)$
= ${\lim }_{x\to 0}\frac{-\sin x-\left(x\cos x+,\sin x\right)+\frac{1}{{\left(1+x\right)}^2}}{2}$
= $\frac{-0-\left(0+0\right)+1}{2}=\frac{1}{2}$