Differentiation Question 480
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
