A) \[{{e}^{-1}}\]
B) \[{{e}^{-4}}\]
C) \[(1+{{e}^{-2}})\]
D) \[{{e}^{-2}}\]
Correct Answer: D
Solution :
[d] Consider \[\underset{x\to \infty }{\mathop{\lim }}\,\left[ \frac{{{x}^{100}}}{{{e}^{x}}}+{{\left( \cos \frac{2}{x} \right)}^{{{x}^{2}}}} \right]\] \[=\underset{x\to \infty }{\mathop{\lim }}\,\frac{{{x}^{100}}}{{{e}^{x}}}+\underset{x\to \infty }{\mathop{\lim }}\,{{\left[ \cos \left( \frac{2}{x} \right) \right]}^{{{x}^{2}}}}\] \[=\underset{x\to \infty }{\mathop{\lim }}\,\frac{{{x}^{100}}}{{{e}^{x}}}=0\,\,\,\,(using\,\,L'Hopsital's\,\,rule)\] and \[\underset{x\to \infty }{\mathop{\lim }}\,{{\left( \cos \frac{2}{x} \right)}^{{{x}^{2}}}}\]is of \[({{1}^{\infty }})\]form \[={{e}^{\underset{x\to \infty }{\mathop{\lim }}\,{{x}^{2}}\left( \cos \frac{2}{x}-1 \right)}}\] \[={{e}^{\underset{t\to 0}{\mathop{\lim }}\,\frac{4}{{{t}^{2}}}(cost-1)}}\] \[\left( Put\frac{2}{x}=t\Rightarrow x=\frac{2}{t} \right)\] \[={{e}^{-\underset{t\to 0}{\mathop{\lim }}\,\left( \frac{1-\cos t}{{{t}^{2}}} \right)\,.\,4}}={{e}^{-\underset{t\to 0}{\mathop{\lim }}\,\left( \frac{\sin t}{2t} \right)4}}={{e}^{-2}}\]
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