A) 0
B) 1
C) 1/2
D) \[-1/2\]
E) \[\infty \]
Correct Answer: C
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x-\sin x}{{{x}^{3}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x(1-\cos x)}{{{x}^{3}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x}{x}.\frac{2{{\sin }^{2}}x/2}{4.{{(x/2)}^{2}}}\] \[=\frac{1}{2}\]You need to login to perform this action.
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