Question

If \[ \lim_{x\to\infty}\left(1+\frac{a}{x}+\frac{b}{x^2}\right)^{2x}=e^2, \] then \(a=\)

• A. \(2\)
• B. \(1\)
• C. \(\frac{1}{2}\)
• D. \(0\)

Hint:

For limits of the form \(\left(1+\frac{f(x)}{x}\right)^{cx}\), only the coefficient of \(\frac1x\) contributes to the exponent of \(e\). Higher-order terms such as \(\frac1{x^2}\) vanish as \(x\to\infty\).

Answer 1

The Correct Option is B

Concept: Use the standard limit \[ \lim_{x\to\infty}\left(1+\frac{k}{x}\right)^x=e^k. \] For limits of the form \[ \left(1+\frac{a}{x}+\frac{b}{x^2}\right)^{2x}, \] the dominant term inside the bracket is \(\dfrac{a}{x}\).

Step 1: Take logarithm of the given limit. Let \[ L=\lim_{x\to\infty}\left(1+\frac{a}{x}+\frac{b}{x^2}\right)^{2x}. \] Then \[ \ln L = \lim_{x\to\infty} 2x\ln\left(1+\frac{a}{x}+\frac{b}{x^2}\right). \] Using \[ \ln(1+t)=t+o(t), \] where \[ t=\frac{a}{x}+\frac{b}{x^2}, \] we get \[ \ln L = \lim_{x\to\infty} 2x\left(\frac{a}{x}+\frac{b}{x^2}\right). \] \[ = \lim_{x\to\infty} \left(2a+\frac{2b}{x}\right). \] \[ =2a. \] Hence, \[ L=e^{2a}. \]

Step 2: Use the given value of the limit. Given \[ L=e^2. \] Therefore, \[ e^{2a}=e^2. \] Comparing exponents, \[ 2a=2. \] \[ a=1. \] Therefore, \[ \boxed{a=1} \] \[ \boxed{\text{Answer = (B)}} \]