Question

The value of $\underset{x\to \infty }{\lim }{\left(\frac{x^2-2x+1}{x^2-4x+2}\right)}^x$ is

• A. $e^2$
• B. $e^{-2}$
• C. $e^6$
• D. None of these

Answer 1

Answer: (A)

$\underset{x\to \infty }{\lim }{\left(\frac{x^2-2x+1}{x^2-4x+2}\right)}^x$
$=\underset{x\to \infty }{\lim }{\left(1+\frac{2x-1}{x^2-4x+2}\right)}^x$
$=e^{\underset{x\to \infty }{\lim }}\left(\frac{x\left(2x-1\right)}{x^2-4x+2}\right)=e^2$