A) 0
B) 1
C) 18
D) 36
Correct Answer: C
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{\sin }^{2}}3x}{{{x}^{2}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,2{{\left( \frac{\sin 3x}{3x} \right)}^{2}}\times \frac{9}{1}\] \[=18\] Alternate Solution: \[\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{\sin }^{2}}3x}{{{x}^{2}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{1-\cos 6x}{{{x}^{2}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{1-\left( 1+\frac{{{(6x)}^{2}}}{2!}+\frac{{{(6x)}^{4}}}{4!}+.... \right)}{{{x}^{2}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}\left( \frac{{{6}^{2}}}{2!}+\frac{{{6}^{4}}{{x}^{2}}}{4!}+.... \right)}{{{x}^{2}}}\] \[=18\]
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