A) 10/3
B) 3/10
C) 6/5
D) 5/6
Correct Answer: A
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{1-(1-2{{\sin }^{2}}x)\sin 5x}{{{x}^{2}}\sin 3x}\] \[\Rightarrow \]\[\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{\sin }^{2}}x\sin 5x.3x.5x}{{{x}^{2}}\sin 3x.3x.5x}\] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}=1=\frac{10}{3}\]You need to login to perform this action.
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