Q. If ${\lim_{x \rightarrow \infty}\left( 1 + \frac{a}{x} + \frac{b}{x^{2}} \right)}^{2x} = e^{2}$, then the values 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) $a = 1,\ b = 2$
Explanation
(b)
${\lim_{x \rightarrow \infty}\left( 1 + \frac{a}{x} + \frac{b}{x^{2}} \right)}^{2x} = {\lim_{x \rightarrow \infty}\left( 1 + \frac{a}{x} + \frac{b}{x^{2}} \right)}^{2x\left( \frac{\frac{a}{x} + \frac{b}{x^{2}}}{\frac{a}{x} + \frac{b}{x^{2}}} \right)}$
$= e^{\lim_{x \rightarrow \infty}2x(\frac{a}{x} + \frac{b}{x^{2}})}\$
$\lbrack\because\lim_{x \rightarrow \infty}\ (1 + x)^{\frac{1}{x}} = e\rbrack$
$\Rightarrow \ \ \ e^{2} = e^{\lim_{x \rightarrow \infty}\ 2(a + b/x)} = e^{2a}$ [given]
$\Rightarrow \ \ \ \ \ a = 1$
and $b \in R$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.