lim x → 0 (e^(2|sin x|)-2|sin x|-1)/x²

Solution

Maths sir · 4 steps

Call t = 2|sin x|. The top becomes eᵗ - t - 1, and that behaves like t²/2 near zero.

1. Substitute. Let t = 2|sin x|. As x → 0, t → 0⁺, and the numerator is eᵗ - t - 1, a 0/0 form.

   eᵗ - t - 1 = t²/2 + t³/6 + … ⇒ lim(t → 0)(eᵗ - t - 1)/t² = 1/2

2. Rewrite the limit. Multiply and divide by t² = 4 sin² x. The limit becomes (eᵗ - t - 1)/t² × (4 sin² x)/x².
3. Absolute value vanishes. |sin x|² = sin² x, so the left-hand and right-hand limits are the same. Also (sin² x)/x² → 1.
4. So. The limit is 1/2 × 4 × 1 = 2, which is option (d).

The idea to keep. Whenever you see eᵗ - t - 1 with t → 0, replace it by t²/2. A modulus inside a squared term does no harm, because |u|² = u².

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