Q. The value of ${\lim_{x \rightarrow \infty}\left( \frac{x^{2} - 2x + 1}{x^{2} - 4x + 2} \right)}^{x}$ is
A) $e^{2}$
B) $e^{- 2}$
C) $e^{6}$
D) None of these
Explanation
(a)
${\lim_{x \rightarrow \infty}\left( \frac{x^{2} - 2x + 1}{x^{2} - 4x + 2} \right)}^{x} = {\lim_{x \rightarrow \infty}\left( 1 + \frac{2x - 1}{x^{2} - 4x + 2} \right)}^{x}$
$= e^{\lim_{x \rightarrow \infty}\left( \frac{x(2x - 1)}{x^{2} - 4x + 2} \right)} = e^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.