A) \[0\]
B) \[1\]
C) \[1/2\]
D) \[-1/2\]
Correct Answer: C
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x-\sin x}{{{x}^{3}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x(1-\cos x)}{{{x}^{3}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x}{x}.\frac{2{{\sin }^{2}}x/2}{4{{(x/2)}^{2}}}=\frac{1}{2}\] Alternative Solution: \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x-\sin x}{{{x}^{3}}}\] \[\left( x+\frac{{{x}^{3}}}{3}+\frac{2{{x}^{5}}}{15}+... \right)\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\left( x-\frac{{{x}^{3}}}{3!}+\frac{{{x}^{5}}}{5!}+... \right)}{{{x}^{3}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{3}}\left( \frac{1}{3}+\frac{1}{3!} \right)+{{x}^{5}}\left( \frac{2}{15}-\frac{1}{5!} \right)+...}{{{x}^{3}}}\] \[=\frac{1}{2}\] Note: L' Hospital's rule can be used if the form is\[\left( \frac{0}{0} \right)\]and\[\left( \frac{\infty }{\infty } \right)\].
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