A) 0
B) e
C) 1/e
D) 1
Correct Answer: D
Solution :
\[\underset{m\to \infty }{\mathop{\lim }}\,{{\left( \cos \frac{x}{m} \right)}^{m}}=\underset{m\to \infty }{\mathop{\lim }}\,{{\left[ 1+\left( \cos \frac{x}{m}-1 \right) \right]}^{m}}\] \[=\underset{m\to \infty }{\mathop{\lim }}\,{{\left[ 1-\left( -\cos \frac{x}{m}+1 \right) \right]}^{m}}\]\[=\underset{m\to \infty }{\mathop{\lim }}\,{{\left[ 1-2{{\sin }^{2}}\frac{x}{2m} \right]}^{m}}\] \[={{e}^{\underset{m\to \infty }{\mathop{\lim }}\,-\left( 2{{\sin }^{2}}\frac{x}{2m} \right)\,m}}\]\[={{e}^{\underset{m\to \infty }{\mathop{\lim }}\,-2{{\left( \frac{\sin \frac{x}{2m}}{x/2m} \right)}^{2}}\left( \frac{{{x}^{2}}}{4{{m}^{2}}} \right)\,m}}\] \[={{e}^{-2\underset{m\to \infty }{\mathop{\lim }}\,\frac{{{x}^{2}}}{4m}}}={{e}^{0}}=1\].
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