Question

If $\underset{x\to \infty }{lim}{\left(\begin{array}{c}1+\frac{a}{x}+\frac{b}{x^2}\end{array}\right)}^{2x}=e^2$, then the values of $a$ and $b$ are -

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

Answer 1

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

$\underset{x\to \infty }{lim}{\left(\begin{array}{c}1+\frac{a}{x}+\frac{b}{x^2}\end{array}\right)}^{2x}=e^2$

$\Rightarrow \underset{e^{x\to \infty }}{lim}2x\left(\begin{array}{c}1+\frac{a}{x}+\frac{b}{x^2}-1\end{array}\right)=e^2$

$\Rightarrow e^{2a}=e^2$

$\Rightarrow a=1\&b\in R$