Question

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

• A. $\frac{1}{2}$
• B. 0
• C. 1
• D. none of these

Answer 1

Answer: (A)

$\frac{1}{2}$

$\underset{x\to 0}{lim}\frac{x\cos x-\log (1+x)}{x^2}$
$=\underset{x\to 0}{lim}\frac{\cos x-x\sin x-\frac{1}{1+x}}{2x}$ (Applying L'Hospital's rule)
$=\underset{x\to 0}{lim}\frac{\cos x+x\cos x-x\sin x-x^2\sin x-1}{2x(1+x)}=\frac{1}{2}\underset{x\to 0}{lim}\frac{1}{(1+x)}\underset{x\to 0}{lim}\frac{\cos x+x\cos x-x\sin x-x^2\sin x-1}{x}$

$=\frac{1}{2}\underset{x\to 0}{lim}\frac{-\sin x-x\sin x+\cos x-x\cos x-x\sin x-x^2\sin x-2x\sin x}{1}$ (Applying L'Hospital's rule)

$=\frac{1}{2}$
Hence, option 'A' is correct.