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

JEE MATHEMATICS MCQ: LIMITS

1 Easy

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Application +4 | -1

The value of

\lim\limits_{x \to \infty } {\left( {\frac{{{x^2} - 2x + 1}}{{{x^2} - 4x + 2}}} \right)^x}

, is

View answer and explanation

Correct option: A

{e^2}

Option A (

{e^2}

) is correct because the limit can be transformed into the form

{e^{\lim f(x)}}

and the exponent evaluates to 2.

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