Question
$$\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{x^{100}}}}{{{e^x}}} + {{\left( {\cos \frac{2}{x}} \right)}^{{x^2}}}} \right) = ?$$
A.
$${e^{ - 1}}$$
B.
$${e^{ - 4}}$$
C.
$$\left( {1 + {e^{ - 2}}} \right)$$
D.
$${e^{ - 2}}$$
Answer :
$${e^{ - 2}}$$
Solution :
$$\eqalign{
& {\text{Consider }}\mathop {\lim }\limits_{x \to \infty } \left[ {\frac{{{x^{100}}}}{{{e^x}}} + {{\left( {\cos \frac{2}{x}} \right)}^{{x^2}}}} \right] \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^{100}}}}{{{e^x}}} + \mathop {\lim }\limits_{x \to \infty } {\left[ {\cos \left( {\frac{2}{x}} \right)} \right]^{{x^2}}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^{100}}}}{{{e^x}}} = 0\,\,\,\left( {{\text{Using L'Hospital's rule}}} \right) \cr
& {\text{and }}\mathop {\lim }\limits_{x \to \infty } {\left( {\cos \frac{2}{x}} \right)^{{x^2}}}{\text{ is of }}\left( {{1^\infty }} \right){\text{ form}} \cr
& = {e^{\mathop {\lim }\limits_{x \to \infty } \,{x^2}\left( {\cos \frac{2}{x} - 1} \right)}}\,\,\,\left( {{\text{Put }}\frac{2}{x} = t\,\, \Rightarrow x = \frac{2}{t}} \right) \cr
& = {e^{\mathop {\lim }\limits_{t \to 0} \,\frac{4}{{{t^2}}}\left( {\cos \,t - 1} \right)}}\, \cr
& = {e^{ - \mathop {\lim }\limits_{t \to 0} \left( {\frac{{1 - \cos \,t}}{{{t^2}}}} \right).4}} \cr
& = {e^{\mathop { - \lim }\limits_{t \to 0} \left( {\frac{{\sin \,t}}{{2t}}} \right)4}} \cr
& = {e^{ - 2}} \cr} $$