Differentiation Question 499
Question: If $ \underset{x\to \infty }{\mathop{\lim }},{{( 1+\frac{a}{x}+\frac{b}{x^{2}} )}^{2x}}=e^{2}, $ then the values of a and b, are
Options:
A) $ a=1 $ and $ b=2 $
B) $ a=1,b\in R $
C) $ a\in R,b=2 $
D) $ a\in R,b\in R $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] We know that $ \underset{x\to \infty }{\mathop{\lim }},{{( 1+x )}^{\frac{1}{x}}}=e $ We have $ \underset{x\to \infty }{\mathop{\lim }},{{( 1+\frac{a}{x}+\frac{b}{x^{2}} )}^{2x}}=e^{2} $
$ \Rightarrow \underset{x\to \infty }{\mathop{\lim }},{{[ {{( 1+\frac{a}{x}+\frac{b}{x^{2}} )}^{( \frac{1}{\frac{a}{x}+\frac{b}{x^{2}}} )}} ]}^{2x( \frac{a}{x}+\frac{b}{x^{2}} )}}=e^{2} $
$ \Rightarrow {e^{\underset{x,\to ,\infty }{\mathop{\lim }},2}}^{[ a+\frac{b}{x} ]}=e^{2}\Rightarrow e^{2a}=e^{2} $
$ \Rightarrow a=1 $ and $ b\in R $
BETA
