Question

The Value of $\mathop {\lim }\limits_{x \to \infty } {\left[ {\dfrac{{\left( {{x^2} - 2x + 1} \right)}}{{\left( {{x^2} - 4x + 2} \right)}}} \right]^x}$ is
1. ${e^2}$
2. ${e^{ - 2}}$
3. ${e^6}$
4. None of these

Answer 1

Hint: In this question, check the given function is in which form by putting the value of $x$ (as in limit) then solve the indeterminate form using the relevant formula. Here the function is in $\mathop {\lim }\limits_{x \to \infty } f{\left( x \right)^{g\left( x \right)}}$ form (indeterminate form ${\infty ^\infty }$). To solve, use $\mathop {\lim }\limits_{x \to \infty } {e^{g\left( x \right)\left[ {f\left( x \right) - 1} \right]}}$ this formula and solve.

Formula Used:
Indeterminate form (${\infty ^\infty }$) formula –
$\mathop {\lim }\limits_{x \to \infty } {e^{g\left( x \right)\left[ {f\left( x \right) - 1} \right]}}$

Complete step by step Solution:
Given that,
$\mathop {\lim }\limits_{x \to \infty } {\left[ {\dfrac{{\left( {{x^2} - 2x + 1} \right)}}{{\left( {{x^2} - 4x + 2} \right)}}} \right]^x}$
This function is in the form of $\mathop {\lim }\limits_{x \to \infty } f{\left( x \right)^{g\left( x \right)}}$( ${\infty ^\infty }$ form).
Where $f\left( x \right) = \dfrac{{\left( {{x^2} - 2x + 1} \right)}}{{\left( {{x^2} - 4x + 2} \right)}},g\left( x \right) = x$
To solve the indeterminate form use $\mathop {\lim }\limits_{x \to \infty } {e^{g\left( x \right)\left[ {f\left( x \right) - 1} \right]}}$,
$ = \mathop {\lim }\limits_{x \to \infty } {e^{g\left( x \right)\left[ {f\left( x \right) - 1} \right]}}$
$ = \mathop {\lim }\limits_{x \to \infty } {e^{x\left[ {\dfrac{{\left( {{x^2} - 2x + 1} \right)}}{{\left( {{x^2} - 4x + 2} \right)}} - 1} \right]}}$
$ = \mathop {\lim }\limits_{x \to \infty } {e^{x\left[ {\dfrac{{\left( {{x^2} - 2x + 2x - 2x + 1 + 1 - 1} \right)}}{{\left( {{x^2} - 4x + 2} \right)}} - 1} \right]}}$
$ = \mathop {\lim }\limits_{x \to \infty } {e^{x\left[ {1 + \dfrac{{\left( {2x - 1} \right)}}{{\left( {{x^2} - 4x + 2} \right)}} - 1} \right]}}$
$ = \mathop {\lim }\limits_{x \to \infty } {e^{\left[ {\dfrac{{\left( {2{x^2} - x} \right)}}{{\left( {{x^2} - 4x + 2} \right)}}} \right]}}$
$ = \mathop {\lim }\limits_{x \to \infty } {e^{\left[ {\dfrac{{{x^2}\left( {2 - \dfrac{1}{x}} \right)}}{{{x^2}\left( {1 - \dfrac{4}{x} + \dfrac{2}{{{x^2}}}} \right)}}} \right]}}$
$ = {e^2}$

Hence, the correct option is 1.

Note: The key concept involved in solving this problem is a good knowledge of forms of limits. Students must remember that some limits are referred to as indeterminate if the limiting behavior of specific portions of a given statement cannot predict the overall limit. When the limits for the given function are applied, it becomes $\dfrac{0}{0}$, which is known as indeterminate forms.