If α, β are the distinct roots of x² + bx + c = 0, then
lim _(x → β) (e^(2(x² + bx + c)) - 1 - 2(x² + bx + c))/(x - β )² is equal to :

Solution

Maths sir · 4 steps

Call the quadratic t; it goes to 0, so use eᵘ - 1 - u ≈ u²/2 and factor t.

1. Factor the quadratic. Since α and β are its roots, t = x² + bx + c = (x-α)(x-β). As x → β, t → 0.
2. Use the standard limit. Put u = 2t. Then lim(u → 0) (eᵘ - 1 - u)/u² = 1/2, so the numerator behaves like (2t)²/2 = 2t².

   lim(x → β) 2(x-α)²(x-β)²/(x-β)² = 2(β - α)²

3. Use the root relations. With α + β = -b and αβ = c, (β - α)² = (α+β)² - 4αβ = b² - 4c.
4. So. The limit equals 2(b² - 4c), which is option (c).

The idea to keep. When u → 0, eᵘ - 1 - u behaves like u²/2. If the quadratic is the u, factor it through its roots so the (x-β) cancels.

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