A) e
B) \[\frac{1}{e}\]
C) 0
D) 1
Correct Answer: D
Solution :
[d]\[\underset{e\to \infty }{\mathop{\lim }}\,{{\left( \cos \frac{x}{n} \right)}^{n}}={{e}^{\underset{e\to \infty }{\mathop{\lim }}\,}}n\left( \cos \frac{x}{n}-1 \right)\] \[={{e}^{\underset{e\to \infty }{\mathop{\lim }}\,-2n{{\sin }^{2}}\frac{x}{2n}}}={{e}^{\underset{n\to \infty }{\mathop{\lim }}\,-2n\left( \frac{\sin \left( \frac{x}{2n} \right)}{\frac{x}{2n}} \right)\times \frac{{{x}^{2}}}{4{{n}^{2}}}}}\] \[={{e}^{\underset{n\to \infty }{\mathop{\lim }}\,\left( -2n \right)\times \frac{{{x}^{2}}}{4n}={{e}^{{}^\circ }}=1}}\] Hence, option [d]is correct,
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