Question

If \(\displaystyle\lim_{x\rightarrow \infty}\left(1+ \dfrac{a}{x}+\dfrac{b}{x^2}\right)^{2x}=e^2\), then the value of a and b are
1. a ∈ R, b = 2
2. a = 1, b ∈ R
3. a ∈ R, b ∈ R
4. None of these

Answer 1

Correct Answer - Option 2 : a = 1, b ∈ R

Concept: 

The limit of indeterminate form \({1}^{\infty}\) is calculated as follows: \(\underset{x\to a}{\mathop{\lim }}\,f{{\left( x \right)}^{g\left( x \right)}}={{e}^{\underset{x\to a}{\mathop{\lim }}\,\left( f\left( x \right)-1 \right)\left( g\left( x \right) \right)}}\)

Calculation: 

\(\begin{align} & \underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\frac{a}{x}+\frac{b}{{{x}^{2}}} \right)}^{2x}}={{e}^{\underset{x\to \infty }{\mathop{\lim }}\,\left( 1+\frac{a}{x}+\frac{b}{{{x}^{2}}}-1 \right)\left( 2x \right)}} \\ & {{e}^{\underset{x\to \infty}{\mathop{\lim }}\,\left( 2a+\frac{2b}{x} \right)}}= {{e}^{2}}\\ \end{align} \)

⇒ e² = e^2a

⇒ 2a = 2

⇒ a = 1 and b can take any real value because any real number divided by infinity is zero. 

Hence, a = 1 and b ∈ R