A) 0
B) 1
C) ½
D) 1/3
Correct Answer: C
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\,\,\,\frac{\tan \,\,2x-x}{3x-\sin x}=\underset{x\to 0}{\mathop{\lim }}\,\,\,\,\left\{ \frac{\frac{2\,\tan 2x}{2x}-1}{3-\frac{\sin x}{x}} \right\}=\frac{1}{2}.\] Aliter : Apply L-Hospital?s rule \[\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{\tan 2x-x}{3x-\sin x}=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{2{{\sec }^{2}}2x-1}{3-\cos x}=\frac{2-1}{3-1}=\frac{1}{2}.\]
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