A) \[\frac{1}{3}\]
B) \[-\frac{1}{3}\]
C) \[\frac{1}{6}\]
D) \[-\frac{1}{6}\]
Correct Answer: D
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\,\frac{\sin x-x}{{{x}^{3}}}\] Expand sin x, then \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{-\frac{{{x}^{3}}}{3\,!}+\frac{{{x}^{5}}}{5\,!}-...}{{{x}^{3}}}=\underset{x\to 0}{\mathop{\lim }}\,\left[ -\frac{1}{3\,!}+\frac{{{x}^{2}}}{5\,!}-... \right]=\frac{-1}{3\,!}=\frac{-1}{6}\]. Aliter : Apply L-Hospital?s rule.
You need to login to perform this action.
You will be redirected in
3 sec
