Functions Question 74
Question: $ \underset{x\to 0}{\mathop{\lim }},\frac{\sin x-x}{x^{3}}= $
[MNR 1980, 86]
Options:
A) $ \frac{1}{3} $
B) $ -\frac{1}{3} $
C) $ \frac{1}{6} $
D) $ -\frac{1}{6} $
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Answer:
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 }},[ -\frac{1}{3,!}+\frac{x^{2}}{5,!}-… ]=\frac{-1}{3,!}=\frac{-1}{6} $ . Aliter : Apply L-Hospital?s rule.
BETA
