JEE Main 2020 · 8 January · morning · Single correct · +4 / −1

lim _(x → 0) ( (3x² + 2)/(7x² + 2) )^(1/x²) is equal to

  1. (a) e
  2. (b) e²
  3. (c) 1/e²
  4. (d) 1/e
Answer and solution

Answer (c) 1/e²

Solution

Maths sir · 4 steps

Base goes to 1 and power to infinity: use eˡⁱᵐᵍ⁽ᶠ⁻¹⁾ and the rest is one line of algebra.

  1. Spot the form. As x → 0, the base (3x²+2)/(7x²+2) → 2/2 = 1 and the power 1/x² → ∞. This is the 1^∞ form.

    lim f(x)ᵍ⁽ˣ⁾ = eˡⁱᵐᵍ⁽ˣ⁾⁽ᶠ⁽ˣ⁾⁻¹⁾ when f → 1, g → ∞

  2. Find f minus 1. f(x) - 1 = (3x²+2-7x²-2)/(7x²+2) = -4x²/(7x²+2).
  3. Multiply by the power. 1/x² × -4x²/(7x²+2) = -4/(7x²+2), and this tends to -4/2 = -2 as x → 0.
  4. So. The limit is e⁻² = 1/e², which is option (c).

The idea to keep. Whenever the base tends to 1 and the power to infinity, write the answer as eˡⁱᵐᵍ⁽ᶠ⁻¹⁾. Always subtract 1 from the base itself, keeping the sign.

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

Study Limits

Asked about 110 times in JEE Main since 2020. This question is from 2020.