Rankwise

limx→0 (aˣ − 1) / x = log a

The limit of (aˣ − 1) / x as x tends to 0 is log a.

Proof

  1. Write aˣ = e^(x log a).
  2. (aˣ − 1)/x = [e^(x log a) − 1]/(x log a) × log a.
  3. The bracket tends to 1 (the eˣ limit with x log a in place of x), so the limit is log a.

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 (aˣ − 1) / x as x tends to 0?

log a.

How do you prove lim x→0 (aˣ − 1) / x = log a?

Write aˣ = e^(x log a). (aˣ − 1)/x = [e^(x log a) − 1]/(x log a) × log a. The bracket tends to 1 (the eˣ limit with x log a in place of x), so the limit is log a.

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.