The value of limx→1(x2−1)sin2(πx)x4−2x3+2x−1 is equal to

x4−2x3+2x−1= (x−1)2 (x2−1) sinπx=sin (π (1−x) =−sin (sinπ (x−1)) limx→1 (x2−1)⋅sin2πx (x2−1) (x−1)2=limx→1sin2 (π (x−1)) (x−1)2 =limx→1sin2 (π (x−1)) (π (x−1))2⋅π2 = 2

The value of $\underset{x \to 1}{l i m} \frac{(x^{2} - 1) s i n^{2} (π x)}{x^{4} - 2 x^{3} + 2 x - 1}$ is equal to

Option 1 - <math> <mrow> <mfrac> <mrow> <msup> <mrow> <mi>π</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mn>6</mn> </mrow> </mfrac> </mrow> </math>

Option 2 - <math> <mrow> <mfrac> <mrow> <msup> <mrow> <mi>π</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> </math>

Option 3 - <math> <mrow> <mfrac> <mrow> <msup> <mrow> <mi>π</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> </math>

Option 4 - 2

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Payal Gupta | Contributor-Level 10

6 months ago

Correct Option - 4

Detailed Solution:

$x^{4} - 2 x^{3} + 2 x - 1 = {(x - 1)}^{2} (x^{2} - 1)$

$s i n π x = s i n (π (1 - x)$

$= - s i n (s i n π (x - 1))$

$\underset{x \to 1}{l i m} \frac{(x^{2} - 1) \cdot s i n^{2} π x}{(x^{2} - 1) {(x - 1)}^{2}} = \underset{x \to 1}{l i m} \frac{s i n^{2} (π (x - 1))}{{(x - 1)}^{2}}$