Question

If $\underset{x\to \infty }{lim}{(1+\frac{a}{x}+\frac{b}{x^2})}^{2x}=e^2,$ then the value of $a$ and $b$ can be

• A. $a=1$
• B. $b=1$
• C. $b=2$
• D. $a=2$

Answer 1

Answer: (A), (B), (C)

$b=2$

Let $L=\underset{x\to \infty }{lim}{(1+\frac{a}{x}+\frac{b}{x^2})}^{2x},\left[1^{\infty }\text{form}\right]$
$=e^{\underset{x\to \infty }{lim}(\frac{a}{x}+\frac{b}{x^2})2x}$
$=e^{\underset{x\to \infty }{lim}(a+\frac{b}{x})2}$
$=e^{2a}$
Given that $L=e^2$
$\Rightarrow e^2=e^{2a}$
$\Rightarrow a=1$
and $b\in R$