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Evaluating an infinity^0 limit using logarithms and L'Hopital's Rule

Evaluate the limit lim⁡x→∞x1/x.\lim_{x \to \infty} x^{1/x}.

This limit has the indeterminate form ∞0\infty^0. Use the substitution y=x1/xy = x^{1/x}, take the natural logarithm of both sides, and apply L'Hopital's Rule to the resulting limit before converting back to find the original limit. What is the value of lim⁡x→∞x1/x\lim_{x \to \infty} x^{1/x}?

Hint

Let y=x1/xy = x^{1/x} and take ln⁡y=ln⁡xx\ln y = \frac{\ln x}{x}. Find lim⁡x→∞ln⁡y\lim_{x\to\infty} \ln y first — this is now a ∞∞\frac{\infty}{\infty} form suitable for L'Hopital's Rule. Once you know this limit, exponentiate to recover lim⁡x→∞y\lim_{x\to\infty} y.

Step-by-step solution

Let y=x1/xy = x^{1/x}. Taking the natural log of both sides gives ln⁡y=1xln⁡x=ln⁡xx\ln y = \frac{1}{x}\ln x = \frac{\ln x}{x}, which as x→∞x \to \infty has the form ∞∞\frac{\infty}{\infty}, so L'Hopital's Rule applies. Differentiating numerator and denominator separately: lim⁡x→∞ln⁡xx=lim⁡x→∞1/x1=lim⁡x→∞1x=0\lim_{x\to\infty} \frac{\ln x}{x} = \lim_{x\to\infty} \frac{1/x}{1} = \lim_{x\to\infty} \frac{1}{x} = 0.

Since lim⁡x→∞ln⁡y=0\lim_{x\to\infty} \ln y = 0, and the natural log function is continuous, we can exponentiate both sides: lim⁡x→∞y=e0=1\lim_{x\to\infty} y = e^{0} = 1. Therefore lim⁡x→∞x1/x=1\lim_{x \to \infty} x^{1/x} = 1.

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