AP Calculus AB and BC

Limit of x^(1/ln x) as x Approaches Infinity

Q: Is the function really constant?

Yes, for x > 1. The identity x = e^{\ln x} gives x^{1/\ln x} = e^{\ln x / \ln x} = e^{1}, with no limiting process needed.

The limit of x raised to the power one over ln x as x approaches infinity is e. Taking logarithms turns the exponent into ln x divided by ln x, which is exactly one at every point, so the expression is constantly e.

\lim_{x \to \infty} x^{1/\ln x} = e

Settled by taking logarithms, which collapses the exponent exactly.

Take logarithms first

The form is $\infty^{0}$ , which is indeterminate, so it needs work. Let $y = x^{1/\ln x}$ and take the natural logarithm of both sides.

\ln y = \frac{1}{\ln x}\cdot \ln x = 1

The two logarithms cancel exactly. So $\ln y = 1$ for every $x > 1$ , which means $y = e$ identically, not merely in the limit.

A limit that was never really a limit

This is the rare case where the function is CONSTANT on its domain. The expression equals $e$ at $x = 2$ , at $x = 10^{6}$ , and everywhere else past 1, so the limit is $e$ for the least interesting reason possible.

The lesson generalises: simplify before taking limits. An $\infty^{0}$ form looks like it needs L'Hopital's rule, and here the algebra removes the problem entirely.

Not sure which technique a limit wants?

The Limit Method Chooser walks the decision from direct substitution through factoring, the conjugate, and L'Hopital, and says why each one applies or fails.

Frequently asked questions

Why is infinity to the zero indeterminate?

Because the base and the exponent pull in opposite directions and the winner depends on their rates. $x^{1/x} \to 1$ , this one gives e, and other pairings give anything at all.

Is the function really constant?

Yes, for $x > 1$ . The identity $x = e^{\ln x}$ gives $x^{1/\ln x} = e^{\ln x / \ln x} = e^{1}$ , with no limiting process needed.