A) \[\frac{1}{2}\]
B) \[-\frac{1}{2}\]
C) 0
D) 1
Correct Answer: A
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,=\frac{\tan x-\sin x}{{{x}^{3}}}\] \[\left( \frac{0}{0}form \right)\] (using L Hospitals rule) \[\underset{x\to 0}{\mathop{\lim }}\,=\frac{{{\sec }^{2}}x-\cos x}{3{{x}^{2}}}\] \[\left( \frac{0}{0}form \right)\] (using L Hospitals rule) \[\underset{x\to 0}{\mathop{\lim }}\,=\frac{2{{\sec }^{2}}x\tan x+\sin x}{6x}\] \[\left( \frac{0}{0}form \right)\] \[\underset{x\to 0}{\mathop{\lim }}\,=\frac{\begin{align} & 2({{\sec }^{2}}x\,.\,\,{{\sec }^{2}}x2\sec x. \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sec x\tan x\,.\,\,\tan x)+\cos x \\ \end{align}}{6}\] (Using L Hospitals rule) \[=\frac{2\,\,(1.1+2\,\,(0))+1}{6}\] \[=\frac{3}{6}=\frac{1}{2}\]
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