Question
If $$\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{a}{x} + \frac{b}{{{x^2}}}} \right)^{2x}} = {e^2},$$ then the value of $$a$$ and $$b,$$ are-
A.
$$a=1$$ and $$b=2$$
B.
$$a = 1,\,\,b \in {\bf{R}}$$
C.
$$a \in {\bf{R}},\,\,b = 2$$
D.
$$a \in {\bf{R}},\,\,b \in {\bf{R}}$$
Answer :
$$a = 1,\,\,b \in {\bf{R}}$$
Solution :
We know that
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } {\left( {1 + x} \right)^{\frac{1}{x}}} = e \cr
& \therefore \,\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{a}{x} + \frac{b}{{{x^2}}}} \right)^{2x}} = {e^2} \cr
& = \mathop {\lim }\limits_{x \to \infty } {\left[ {\left( {1 + \frac{a}{x} + \frac{b}{{{x^2}}}} \right)\left( {\frac{1}{{\frac{a}{x} + \frac{b}{{{x^2}}}}}} \right)} \right]^{2x\left( {\frac{a}{x} + \frac{b}{{{x^2}}}} \right)}} = {e^2} \cr
& = {e^{\mathop {\lim }\limits_{x \to \infty } 2\left[ {a + \frac{b}{x}} \right]}} \cr
& = {e^2} \cr
& \Rightarrow {e^{2a}} = {e^2} \cr
& \Rightarrow a = 1{\text{ and }}b \in {\bf{R}} \cr} $$