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JEE Main 202229 JuneEvening Shift
Exam
JEE Main
Year
2022
Date
29 June
Shift
Evening Shift
Subject
Mathematics
Chapter
Limits

JEE Main 2022 · 29 June · Evening Shift · Mathematics · Limits

The value of _ x → 1 ( x ^ 2 - 1) ^ 2 ( ) x ^ 4 - 2 x ^ 3 + 2x - 1...

The value of lim⁡x→1(x2−1)sin⁡2(\pix)x4−2x3+2x−1\lim_{x \to 1}\dfrac{\left({x}^{2} - 1\right)\sin^{2}\left(\pix\right)}{{x}^{4} - 2{x}^{3} + 2x - 1} is equal to:

Options

  1. Aπ26\dfrac{{\pi}^{2}}{6}
  2. Bπ23\dfrac{{\pi}^{2}}{3}
  3. Cπ22\dfrac{{\pi}^{2}}{2}
  4. Dπ2{\pi}^{2}

Correct answer

D. π2{\pi}^{2}

Step-by-step solution

Given, lim⁡x→1(x2−1)sin⁡2(\pix)x4−2x3+2x−1\lim_{x \to 1}\dfrac{\left({x}^{2} - 1\right)\sin^{2}\left(\pix\right)}{{x}^{4} - 2{x}^{3} + 2x - 1}

=lim⁡x→1(x2−1)sin⁡2(\pix)(x2−1)(x2−2x+1)= \lim_{x \to 1}\dfrac{\left({x}^{2} - 1\right)\sin^{2}\left(\pix\right)}{\left({x}^{2} - 1\right)\left({x}^{2} - 2x + 1\right)}

=lim⁡h→0sin⁡2(π(1+h))(1+h−1)2= \lim_{h \to 0}\dfrac{\sin^{2}\left(\pi\left(1 + h\right)\right)}{{\left(1 + h - 1\right)}^{2}}

=lim⁡h→0sin⁡2\pihh2=lim⁡h→0(sin\pih\pih)2π2=π2= \lim_{h \to 0}\dfrac{\sin^{2}\pih}{{h}^{2}} = \lim_{h \to 0}{\left(\dfrac{sin\pih}{\pih}\right)}^{2}{\pi}^{2} = {\pi}^{2}

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