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lim_x→0⁺ xˣ = 1

The limit of xˣ as x tends to 0⁺ is 1.

Proof

Let y = xˣ, so log y = x log x.
x log x → 0 as x → 0⁺, so log y → 0 and y → e⁰ = 1.

Standard limits

Limit	Value
lim_x→0 sin x / x	1
lim_x→0 tan x / x	1
lim_x→0 (1 − cos x) / x²	1/2
lim_x→0 (1 − cos x) / x	0
lim_x→0 (eˣ − 1) / x	1
lim_x→0 (aˣ − 1) / x	log a
lim_x→0 log(1 + x) / x	1
lim_x→0 (1 + x)^(1/x)	e
lim_x→∞ (1 + 1/x)ˣ	e
lim_x→∞ (1 + a/x)ˣ	eᵃ
lim_x→a (xⁿ − aⁿ) / (x − a)	n aⁿ⁻¹
lim_x→0 sin⁻¹x / x	1
lim_x→0 tan⁻¹x / x	1
lim_x→∞ sin x / x	0
lim_x→0⁺ x log x	0
lim_x→0⁺ xˣ	1

Questions students ask

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

1.

How do you prove lim x→0⁺ xˣ = 1?

Let y = xˣ, so log y = x log x. x log x → 0 as x → 0⁺, so log y → 0 and y → e⁰ = 1.

Use these next