The value of $\mathop {\lim }\limits _{x \to 1} \frac{\left(x^2-1\right) \sin ^2(\pi x)}{x^4-2 x^3+2 x-1}$ is equal to

JEE MATHEMATICS MCQ: LIMITS

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The value of

\lim\limits _{x \to 1} \frac{\left(x^2-1\right) \sin ^2(\pi x)}{x^4-2 x^3+2 x-1}

is equal to

View answer and explanation

Correct option: D

\pi^2

Option D (

\pi^2

) is correct because after simplification and applying L'Hopital's rule or trigonometric limits, the limit evaluates to

\pi^2

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