Question

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

• A. $a\in R,b\in R$
• B. $a=1,b\in R$
• C. $a\in R,b=2$
• D. $a=1,b=2$

Answer 1

The correct option is B $a=1,b\in R$

Given, $\underset{x\to \infty }{lim}{(1+\frac{a}{x}+\frac{b}{x^2})}^{2x}=e^2$ ....($1^{\infty }$ form)
$\Rightarrow e^{\underset{x\to \infty }{lim}2x(\frac{a}{x}+\frac{b}{x^2})}=e^2$
$\Rightarrow \underset{x\to \infty }{lim}{e}^{2\left(a+b/x\right)}=e^2$
$\Rightarrow e^{2a}=e^2$
$\Rightarrow 2a=2$
$\Rightarrow a=1$ and $b\in R$