A) 10/3
B) 3/10
C) 6/5
D) 5/6
Correct Answer: A
Solution :
\[\underset{x\to 0}{\mathop{\text{lim}}}\,\frac{(1-\cos 2x)\,\sin 5x}{{{x}^{2}}\sin 3x}\]\[=\underset{x\to 0}{\mathop{\text{lim}}}\,\frac{2{{\sin }^{2}}x\,\sin 5x}{{{x}^{2}}\sin 3x}\] \[=\underset{x\to 0}{\mathop{\text{lim}}}\,\,\left( \frac{2{{\sin }^{2}}x}{{{x}^{2}}} \right)\frac{\left( \frac{\sin 5x}{x} \right)}{\left( \frac{\sin 3x}{x} \right)}\] \[=\underset{x\to 0}{\mathop{\text{lim}}}\,2\,{{\left( \frac{\sin x}{x} \right)}^{2}}\times \frac{5\underset{x\to 0}{\mathop{\text{lim}}}\,\left( \frac{\sin 5x}{5x} \right)}{3\underset{x\to 0}{\mathop{\text{lim}}}\,\left( \frac{\sin 3x}{3x} \right)}\]\[=\frac{2\times 5}{3}=\frac{10}{3}\].
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