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Question

Mathematics Question on Limits

The value of
lim⁡x→1(x2−1)sin⁡2(πx)x4−2x3+2x−1\lim_{{x \to 1}} \frac{{(x^2 - 1) \sin^2(\pi x)}}{{x^4 - 2x^3 + 2x - 1}}
is equal to

A

π26\frac{π²}{6}

B

π23\frac{π²}{3}

C

π22\frac{π²}{2}

D

π²

Answer

π²

Explanation

Solution

The correct answer is (D) : π²
lim⁡x→1(x2−1)sin⁡2(πx)x4−2x3+2x−1\lim_{{x \to 1}} \frac{{(x^2 - 1) \sin^2(\pi x)}}{{x^4 - 2x^3 + 2x - 1}}
=lim⁡x→1(x+1)(x−1)sin⁡2(πx)(x−1)3(x+1)=\lim_{{x \to 1}} \frac{{(x+1)(x-1) \sin^2(\pi x)}}{{(x-1)^3(x+1)}}
Let x-1 = t
lim⁡t→0(2+t)tsin⁡2(πt)t3(t+2)\lim_{{t \to 0}} \frac{{(2+t)t \sin^2(\pi t)}}{{t^3(t+2)}} == lim⁡t→0sin⁡2(πt)π2t2⋅π2\lim_{{t \to 0}} \frac{{\sin^2(\pi t)}}{{\pi^2t^2}} \cdot \pi^2 = π2\pi^2