Rankwise

limx→0⁺ x log x = 0

The limit of x log x as x tends to 0⁺ is 0.

Proof

  1. Write x log x = log x ÷ (1/x), an ∞/∞ form.
  2. L'Hôpital: (1/x) ÷ (−1/x²) = −x.
  3. As x → 0⁺, −x → 0.

Standard limits

LimitValue
limx→0 sin x / x1
limx→0 tan x / x1
limx→0 (1 − cos x) / x²1/2
limx→0 (1 − cos x) / x0
limx→0 (eˣ − 1) / x1
limx→0 (aˣ − 1) / xlog a
limx→0 log(1 + x) / x1
limx→0 (1 + x)^(1/x)e
limx→∞ (1 + 1/x)ˣe
limx→∞ (1 + a/x)ˣeᵃ
limx→a (xⁿ − aⁿ) / (x − a)n aⁿ⁻¹
limx→0 sin⁻¹x / x1
limx→0 tan⁻¹x / x1
limx→∞ sin x / x0
limx→0⁺ x log x0
limx→0⁺ xˣ1

Questions students ask

What is the limit of x log x as x tends to 0⁺?

0.

How do you prove lim x→0⁺ x log x = 0?

Write x log x = log x ÷ (1/x), an ∞/∞ form. L'Hôpital: (1/x) ÷ (−1/x²) = −x. As x → 0⁺, −x → 0.

Use these next

Standard limitssin x/x, (eˣ − 1)/x, (1 + x)^(1/x) and every other standard limit, with proofs.Derivative formulasEvery standard derivative with its proof.Integration formulasEvery standard integral with a step-by-step proof.Maths formula sheetAlgebra to integrals, the formulas JEE Main uses most.