Question

If $\displaystyle\lim_{x \to\infty} \left(1+ \frac{a}{x} + \frac{b}{x^{2}}\right)^{2x} = e^{2}$ , then the values of a and b, are

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

Answer 1

Answer: (B)

$a = 1, b \in R$

Answer 2

We know that $\displaystyle\lim_{x\to\infty} \left(1+x\right)^{\frac{1}{x}} =e$ $\therefore \displaystyle\lim_{x\to\infty} \left(1+ \frac{a}{x} + \frac{b}{x^{2}}\right)^{2x} = e^{2}$ $\Rightarrow \displaystyle\lim_{x\to\infty} \left[\left(1+ \frac{a}{x} + \frac{b}{x^{2}}\right) ^{\left(\frac{1}{\frac{a}{x}+ \frac{b}{x^{2}}}\right)}\right]^{2x\left(\frac{a}{x} + \frac{b}{x^{2}}\right)} =e^{2}$ $\Rightarrow e^{\displaystyle\lim_{x\to\infty}2\left[a+ \frac{b}{x}\right]} = e^{2}$ $\Rightarrow e^{2a} = e^{2}$ $\Rightarrow a=1$ , $b \in R$