A) 1
B) \[e\]
C) \[{{e}^{-1}}\]
D) none of these
Correct Answer: A
Solution :
[a] \[\underset{m\to \infty }{\mathop{\lim }}\,{{\left( \cos \frac{x}{m} \right)}^{m}}=\underset{m\to \infty }{\mathop{\lim }}\,{{\left[ 1-\left( 1-\cos \frac{x}{m} \right) \right]}^{m}}\] \[\underset{m\to \infty }{\mathop{\lim }}\,{{\left[ 1-2{{\sin }^{2}}\frac{x}{2m} \right]}^{m}}\] \[=\underset{{{e}^{m\to \infty }}}{\mathop{\lim }}\,\left( -2{{\sin }^{2}}\frac{x}{2m} \right)m=1\]
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