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Evaluating lim x->0+ x^x using logarithms and L'Hopital's Rule

Evaluate the limit lim⁡x→0+xx.\lim_{x \to 0^+} x^x.

Since direct substitution gives the indeterminate form 000^0, rewrite xxx^x as an exponential: xx=exln⁡xx^x = e^{x\ln x}. First find lim⁡x→0+xln⁡x\lim_{x\to 0^+} x\ln x by converting it into a quotient suitable for L'Hopital's Rule, then use that result to determine the original limit. State the value of lim⁡x→0+xx\lim_{x\to 0^+} x^x.

Hint

Write xln⁡x=ln⁡x1/xx\ln x = \dfrac{\ln x}{1/x}, which has the form −∞/∞-\infty/\infty as x→0+x\to 0^+. Apply L'Hopital's Rule to this quotient, then exponentiate the result since xx=exln⁡xx^x = e^{x\ln x}.

Step-by-step solution

Rewrite xln⁡xx\ln x as the quotient ln⁡x1/x\dfrac{\ln x}{1/x}. As x→0+x\to 0^+, ln⁡x→−∞\ln x \to -\infty and 1/x→∞1/x \to \infty, so this is the indeterminate form −∞/∞-\infty/\infty, and L'Hopital's Rule applies. Differentiating numerator and denominator separately gives ddxln⁡x=1/x\dfrac{d}{dx}\ln x = 1/x and ddx(1/x)=−1/x2\dfrac{d}{dx}(1/x) = -1/x^2, so the limit becomes lim⁡x→0+1/x−1/x2=lim⁡x→0+(−x)=0\lim_{x\to 0^+} \dfrac{1/x}{-1/x^2} = \lim_{x\to 0^+} (-x) = 0.

Thus lim⁡x→0+xln⁡x=0\lim_{x\to 0^+} x\ln x = 0. Since xx=exln⁡xx^x = e^{x\ln x} and the exponential function is continuous, lim⁡x→0+xx=elim⁡x→0+xln⁡x=e0=1.\lim_{x\to 0^+} x^x = e^{\lim_{x\to 0^+} x\ln x} = e^0 = 1. Therefore the original limit equals 11.

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