Limit

Calculus Level 2

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

− ∞ -\infty Limit does not exist e e 1 1 0 0 ∞ \infty

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Daniel Like by Daniel Like , 115, USA

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1 solution

lim ⁡ x → 0 + x x = lim ⁡ x → 0 + e x ln ⁡ x = lim ⁡ x → 0 + exp ⁡ ( x ln ⁡ x ) exp ⁡ ( x ) = e x = lim ⁡ x → 0 + exp ⁡ ( ln ⁡ x 1 x ) A ∞ / ∞ cases, L’H o ˆ pital’s rule applies. = lim ⁡ x → 0 + exp ⁡ ( 1 x − 1 x 2 ) Differentiating up and down w.r.t. x = lim ⁡ x → 0 + exp ⁡ ( − x ) = e 0 = 1 \begin{aligned} \lim_{x \to 0^+} x^x & = \lim_{x \to 0^+} e^{x\ln x} \\ & = \lim_{x \to 0^+} \color{#3D99F6}{\exp (x\ln x)} \quad \quad \small \color{#3D99F6}{\exp (x) = e^x} \\ & = \lim_{x \to 0^+} \exp \color{#3D99F6}{\left(\frac {\ln x}{\frac 1x} \right)} \quad \quad \small \color{#3D99F6}{\text{A }\infty / \infty \text{ cases, L'Hôpital's rule applies.}} \\ & = \lim_{x \to 0^+} \exp \color{#3D99F6}{\left(\frac {\frac 1x}{-\frac 1{x^2}} \right)} \quad \quad \small \color{#3D99F6}{\text{Differentiating up and down w.r.t. }x} \\ & = \lim_{x \to 0^+} \exp \left(-x \right) \\ & = e^0 = \boxed{1} \end{aligned}

11 Helpful 0 Interesting 0 Brilliant 0 Confused

I think this is only a one sided limit. $${\frac{-1}_{2}}^{-1/2}$$ Is not on the real axis; any even number less than zero is not on the real axis; therefore this function is not continuous at zero.

Charles White - 4 years, 11 months ago

Sorry I typed this on my iPad and couldn't preview before posting. This ${\frac{-1}_{2}}^{-1/2}$ meant (-1/2)^(-1/2).

Charles White - 4 years, 11 months ago

Yes lnx is not defined in the left neighborhood of 0.

Aditya Chauhan - 4 years, 11 months ago

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