MCQ
$\mathop {\lim }\limits_{x \to 0} \,\frac{{(1 - \cos 2x)\sin 5x}}{{{x^2}\sin 3x}} = . . .$
- ✓$10/3$
- B$3/10$
- C$6/5$
- D$5/6$
$ = \mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{2{{\sin }^2}x\,\sin 5x}}{{{x^2}\sin 3x}}$
$ = \mathop {{\rm{lim}}}\limits_{x \to 0} \,\left( {\frac{{2{{\sin }^2}x}}{{{x^2}}}} \right)\frac{{\left( {\frac{{\sin 5x}}{x}} \right)}}{{\left( {\frac{{\sin 3x}}{x}} \right)}}$
$ = \mathop {{\rm{lim}}}\limits_{x \to 0} 2\,{\left( {\frac{{\sin x}}{x}} \right)^2} \times \frac{{5\mathop {{\rm{lim}}}\limits_{x \to 0} \left( {\frac{{\sin 5x}}{{5x}}} \right)}}{{3\mathop {{\rm{lim}}}\limits_{x \to 0} \left( {\frac{{\sin 3x}}{{3x}}} \right)}}$
$ = \frac{{2 \times 5}}{3} = \frac{{10}}{3}$.
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