AP Calculus AB and BC

Limit of x^(1/x) at Infinity Is 1

Q: Is this the same as the limit of the nth root of n?

Yes, restricted to whole numbers, since \sqrt[n]{n} = n^{1/n}. The sequence tends to 1 for the same logarithm reason, and that fact is what makes the root test usable on series whose terms carry an n out front.

Q: Where is x^(1/x) largest?

At x = e, with value e^{1/e} \approx 1.4447. To the left of e the function rises from 1 at x = 1, and to the right it falls back toward 1 forever without reaching it, since x^{1/x} > 1 for every x > 1.

The limit of x to the power 1/x as x approaches infinity is 1. Substitution gives infinity to the 0, an indeterminate form. Take the natural log: the log of the expression is (ln x)/x, which goes to 0 by L'Hopital's rule, so the expression itself goes to e to the 0, which is 1.

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

Settled by logarithms then L'Hopital.

Logging first, exponentiating last

The exponent is where the trouble is, so move it. Set $y = x^{1/x}$ and take the natural log of both sides, which brings the exponent down in front.

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

That quotient is $\frac{\infty}{\infty}$ , one of the two forms L'Hopital's rule accepts, and a single pass finishes it.

\lim_{x \to \infty} \ln y = \lim_{x \to \infty} \frac{\ln x}{x} = \lim_{x \to \infty} \frac{\frac{1}{x}}{1} = 0

That is the limit of $\ln y$ , not of $y$ . The exponential is continuous, so $y = e^{\ln y}$ follows its logarithm up to $e^{0}$ .

\lim_{x \to \infty} x^{1/x} = e^{0} = 1

$x$	$x^{1/x}$
$10$	$1.25893$
$100$	$1.04713$
$1000$	$1.00693$
$1{,}000{,}000$	$1.0000138$

The graph climbs before it settles

$x^{1/x}$ does not fall the whole way. Its logarithm $\frac{\ln x}{x}$ has derivative $\frac{1-\ln x}{x^2}$ , which is $0$ at $x = e$ , so the maximum sits there with value $e^{1/e} \approx 1.4447$ . That single fact explains why $\sqrt[3]{3} \approx 1.4422$ is larger than $\sqrt{2} \approx 1.4142$ : the cube root of $3$ is closer to the peak.

Why substituting infinity says nothing

The base grows without bound while the exponent shrinks to $0$ , so the expression reads as $\infty^{0}$ .

x^{1/x} \longrightarrow \infty^{0}

That form is indeterminate, and not on a technicality. A huge base pulls the value up, a tiny exponent pulls it down toward $1$ , and which pull wins depends entirely on the two speeds involved.

Two functions with the identical form disagree completely: $\left(e^{x}\right)^{1/x}$ equals $e$ at every $x$ , while $\left(e^{x^{2}}\right)^{1/x}$ equals $e^{x}$ and runs to $\infty$ . Both have a base going to $\infty$ and an exponent going to $0$ .

Logarithms are the standard opening for every power-form indeterminate, $\infty^{0}$ , $0^{0}$ and $1^{\infty}$ alike, because taking $\ln$ turns the exponent into a factor and a product can be rearranged into the quotient L'Hopital's rule needs.

The mistakes students make

Answering $\infty$ from the base or $0$ from the exponent. Neither piece gets to decide on its own.
Answering $1$ because anything to the zero power is $1$ . The number is right and the reasoning is not: the exponent is never actually $0$ , and $\left(e^{x^{2}}\right)^{1/x}$ has the same form with limit $\infty$ .
Stopping at $0$ . The $0$ belongs to $\ln y$ , and every logarithm-first limit ends with an exponentiation step.
Applying L'Hopital's rule to $x^{1/x}$ as written. The rule handles quotients in $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form, and taking the logarithm is what produces one.
Reading $x^{1/x}$ as $\frac{x^{1}}{x}$ or as $\left(\frac{1}{x}\right)^{x}$ . Those are different functions, with limits $1$ and $0$ .

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

Is this the same as the limit of the nth root of n?

Yes, restricted to whole numbers, since $\sqrt[n]{n} = n^{1/n}$ . The sequence tends to $1$ for the same logarithm reason, and that fact is what makes the root test usable on series whose terms carry an $n$ out front.

Why is L'Hopital's rule needed if the exponent already goes to 0?

Because the base is not staying put. The rule that anything to the zero power is $1$ needs a fixed base, and here the base is running to $\infty$ while the exponent shrinks. The logarithm turns that race into $\frac{\ln x}{x}$ , where the speeds can actually be compared.

Where is x^(1/x) largest?

At $x = e$ , with value $e^{1/e} \approx 1.4447$ . To the left of $e$ the function rises from $1$ at $x = 1$ , and to the right it falls back toward $1$ forever without reaching it, since $x^{1/x} > 1$ for every $x > 1$ .