Differentiation Question 166
Question: $ \underset{x\to \infty }{\mathop{\lim }}( \frac{x^{100}}{e^{x}}+{{( \cos \frac{2}{x} )}^{x^{2}}} )= $
Options:
A) $ {e^{-1}} $
B) $ {e^{-4}} $
C) $ (1+{e^{-2}}) $
D) $ {e^{-2}} $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Consider $ \underset{x\to \infty }{\mathop{\lim }}[ \frac{x^{100}}{e^{x}}+{{( \cos \frac{2}{x} )}^{x^{2}}} ] $
$ =\underset{x\to \infty }{\mathop{\lim }}\frac{x^{100}}{e^{x}}+\underset{x\to \infty }{\mathop{\lim }}{{[ \cos ( \frac{2}{x} ) ]}^{x^{2}}} $
$ =\underset{x\to \infty }{\mathop{\lim }}\frac{x^{100}}{e^{x}}=0(usingL’Hopsital’srule) $ and $ \underset{x\to \infty }{\mathop{\lim }}{{( \cos \frac{2}{x} )}^{x^{2}}} $ is of $ ({1^{\infty }}) $ form $ ={e^{\underset{x\to \infty }{\mathop{\lim }}x^{2}( \cos \frac{2}{x}-1 )}} $
$ ={e^{\underset{t\to 0}{\mathop{\lim }}\frac{4}{t^{2}}(cost-1)}} $
$ ( Put\frac{2}{x}=t\Rightarrow x=\frac{2}{t} ) $
$ ={e^{-\underset{t\to 0}{\mathop{\lim }}( \frac{1-\cos t}{t^{2}} ).4}}={e^{-\underset{t\to 0}{\mathop{\lim }}( \frac{\sin t}{2t} )4}}={e^{-2}} $
BETA
