The value of lim _(x → 1) ((x² - 1)sin ²(π x))/(x⁴ - 2x³ + 2x - 1) is equal to:

Solution

Maths sir · 4 steps

Factor the denominator fully, cancel what matches, and only (sin² π x)/(x-1)² is left.

1. Factor the denominator. At x = 1 the denominator is 1 - 2 + 2 - 1 = 0, and its derivative 4x³ - 6x² + 2 is also 0. So (x-1) is a repeated factor. Dividing it out gives x⁴ - 2x³ + 2x - 1 = (x-1)³(x+1).
2. Cancel common factors. The numerator is (x-1)(x+1)sin²(π x). Cancel (x-1)(x+1) from the top and the bottom. Only (x-1)² stays in the denominator.

   lim(x → 1) sin²(π x)/(x-1)² = lim(h → 0) sin²(π + π h)/h² = lim(h → 0) ((sin π h)/(π h))² π²

3. Use the standard limit. Put x = 1 + h. Then sin(π + π h) = -sin(π h), and the minus sign disappears on squaring. Since (sin π h)/(π h) → 1 as h → 0, the limit is π².
4. So. The limit equals π², which is option (d).

The idea to keep. When a polynomial denominator vanishes at the limit point, factor it completely and cancel first. Then shift the variable so you can use (sin θ)/θ → 1 near 0.

Answer from the NTA final answer key. Solution by PrepWiser.