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

The correct option is A

$e^2$

Explanation for the correct option

Given, $\underset{x\to \infty }{\lim }{\left[\frac{x^2-2x+1}{x^2-4x+2}\right]}^x$

$\begin{array}{rcl}& & \begin{array}{cc}\underset{x\to \infty }{\lim }{\left[\frac{x^2-2x+1}{x^2-4x+2}\right]}^x=\underset{x\to \infty }{\lim }{\left(\frac{x^2-2x-2x+2x+1-1+1}{x^2-4x+2}\right)}^x& \\ =\underset{x\to \infty }{\lim }{\left(\frac{\left(x^2-4x+2\right)+2x-1}{x^2-4x+2}\right)}^x& \\ =\underset{x\to \infty }{\lim }{\left(1+\frac{2x-1}{x^2-4x+2}\right)}^x& \\ =\underset{x\to \infty }{\lim }{\left(1+\frac{2x-1}{x^2-4x+2}\right)}^{x\left(\frac{x^2-4x+2}{2x-1}\right)\left(\frac{2x-1}{x^2-4x+2}\right)}& \\ =\underset{x\to \infty }{\lim }{\left[{\left(1+\frac{2x-1}{x^2-4x+2}\right)}^{\left(\frac{x^2-4x+2}{2x-1}\right)}\right]}^{\left(\frac{2x-1}{x^2-4x+2}\right)x}& \\ =\underset{x\to \infty }{\lim }{\left[{\left(1+\frac{a}{n}\right)}^{\left(\frac{n}{a}\right)}\right]}^{\cancel{x^2}\left[\frac{2-\frac{1}{x}}{\cancel{x^2}\left(1-{\displaystyle \frac{4}{x}}+{\displaystyle \frac{2}{x^2}}\right)}\right]}& \left[a=2x-1,n=x^2-4x+2\right]\\ ={\left(e^a\right)}^{\frac{1}{a}\left[\frac{2-\frac{1}{\infty }}{\left(1-{\displaystyle \frac{4}{\infty }}+{\displaystyle \frac{2}{{\infty }^2}}\right)}\right]}& [\because \underset{n\to \infty }{lim}{\left(1+\frac{x}{n}\right)}^n=e^x]\\ ={\left(e\right)}^{a\times \frac{1}{a}\left[\frac{2-0}{\left(1-{\displaystyle 0}+{\displaystyle 0}\right)}\right]}& [\because {\left(K^m\right)}^n={\left(K\right)}^{m\times n},\frac{1}{\infty }=0]\\ =e^{\left[\frac{2}{1}\right]}& \\ =e^2& \end{array}\end{array}$

Therefore, $\underset{x\to \infty }{\lim }{\left[\frac{x^2-2x+1}{x^2-4x+2}\right]}^x=e^2$.

Hence, option(A) is correct.