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