Zero Power Zero

Calculus Level 1

lim ⁡ x → 0 + x x = ? \large \lim_{x\to 0^+} x^x =\, ?


The answer is 1.

Worranat Pakornrat by Worranat Pakornrat , 36, Thailand

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1 solution

L = lim ⁡ x → 0 + x x = lim ⁡ x → 0 + exp ⁡ ( x ln ⁡ x ) = exp ⁡ ( lim ⁡ x → 0 + x ln ⁡ x ) = exp ⁡ ( lim ⁡ x → 0 + ln ⁡ x 1 x ) A ∞ / ∞ case, L’H o ˆ pital’s rule applies. = exp ⁡ ( lim ⁡ x → 0 + 1 x − 1 x 2 ) Differentiate up and down with x = exp ⁡ ( − lim ⁡ x → 0 + 1 x × x 2 ) = e 0 = 1 \begin{aligned} L & = \lim_{x \to 0^+} x^x \\ &= \lim_{x \to 0^+} \exp (x\ln x) \\ & = \exp \left( \lim_{x \to 0^+} x\ln x \right) \\ & = \exp \left( \lim_{x \to 0^+} \frac {\ln x}{\frac 1x} \right) & \small \color{#3D99F6} \text{A }\infty /\infty \text { case, L'Hôpital's rule applies.} \\ & = \exp \left( \lim_{x \to 0^+} \frac {\frac 1x}{-\frac 1{x^2}} \right) & \small \color{#3D99F6} \text{Differentiate up and down with }x \\ & = \exp \left(- \lim_{x \to 0^+} \frac 1x \times x^2 \right) \\ &= e^0 = \boxed{1} \end{aligned}

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