Question

$\underset{x\to 0}{lim}\frac{x\cos x-\sin x}{x^2\sin x}$ equals

• A. $\frac{1}{3}$
• B. $-\frac{1}{3}$
• C. 3
• D. -3

Answer 1

Answer: (B)

$-\frac{1}{3}$

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