AP Calculus AB and BC

Limit of x^x as x Approaches 0 from the Right

Q: Is 0^0 equal to 1?

As a limit form it has no fixed value, which is exactly why \lim_{x \to 0^+} x^x takes work. The answer here is 1, but that is a fact about x^x, not about the symbol 0^0. Other functions with the same form converge to other numbers, and some run off to \infty.

Q: Why is there no left-hand limit?

x^x is really e^{x \ln x}, and \ln x needs x > 0. Negative bases produce real values only at isolated exponents such as -\frac{1}{3}, so there is no interval to the left of 0 on which the function exists.

Q: Can I use L'Hopital directly on x^x?

No. The rule applies to quotients in \frac{0}{0} or \frac{\infty}{\infty} form, and x^x is neither. Taking the logarithm gives the product x \ln x, and rewriting that as \frac{\ln x}{\frac{1}{x}} makes the rule legal.

The limit of x to the x as x approaches 0 from the right is 1. Substitution gives the indeterminate form 0 to the 0, so take the natural log: x ln x goes to 0, and exponentiating back gives e to the 0, which is 1. Only the right-hand limit exists, because x to the x has no real values on any interval to the left of 0.

\lim_{x \to 0^+} x^x = 1

Settled by logarithms then L'Hopital.

Taking the logarithm to unlock the exponent

The exponent is the problem, so move it. Set $y = x^x$ and take the natural log of both sides, which turns the exponent into a factor.

\ln y = x \ln x

The right side is a $0 \cdot (-\infty)$ product, indeterminate on its own. Write it as a quotient so L'Hopital's rule applies.

\lim_{x \to 0^+} x \ln x = \lim_{x \to 0^+} \frac{\ln x}{\frac{1}{x}} = \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{1}{x^2}} = \lim_{x \to 0^+} (-x) = 0

That is the limit of $\ln y$ , not of $y$ . Since $\ln y \to 0$ and the exponential is continuous, $y = e^{\ln y}$ approaches $e^{0}$ .

\lim_{x \to 0^+} x^x = e^{0} = 1

Check it numerically

$0.1^{0.1} \approx 0.7943$ , $0.01^{0.01} \approx 0.9550$ , and $0.001^{0.001} \approx 0.9931$ . The values climb toward 1 slowly, which is why a table alone can leave you unsure. The graph even dips first: $x^x$ bottoms out at $x = \frac{1}{e} \approx 0.3679$ with value about $0.6922$ , then turns back up.

Why substituting x = 0 gives you nothing

Substitution produces $0^0$ , which is an indeterminate form rather than a value. The base is shrinking to 0, which drags the expression toward 0, while the exponent is shrinking to 0, which drags it toward 1. The form on its own does not say which pull wins.

How undecided that form really is: $\left(e^{-\frac{1}{x}}\right)^{x}$ has a base going to 0 and an exponent going to 0 as $x \to 0^+$ , yet it equals $e^{-1} \approx 0.3679$ at every $x$ . Same form, different answer. Only the specific functions decide.

Logarithms are the standard opening for every power-form indeterminate, $0^0$ , $\infty^0$ and $1^\infty$ alike, because taking $\ln$ converts the exponent into a product, and a product can be rewritten as the quotient that L'Hopital's rule (Topic 4.7) needs.

The mistake students make

The costly one is stopping a step early.

Answering 0. The limit of $\ln y$ is 0, so the limit of $y$ is $e^{0} = 1$ . Every logarithm-first limit ends with an exponentiation step, and skipping it is the most common slip on this problem.
Claiming $0^0 = 1$ by the rule that anything to the zero power is 1. That rule needs a fixed nonzero base, and here the base is moving.
Writing a two-sided limit. For $x < 0$ the expression $x^x$ is not real at most points, since $(-0.5)^{-0.5} = \frac{1}{\sqrt{-0.5}}$ is imaginary, so there is no left side to approach from and only $\lim_{x \to 0^+}$ is defined.
Applying L'Hopital to $x \ln x$ as it stands. The rule needs a quotient in $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form, so the product has to be rearranged first.

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 0^0 equal to 1?

As a limit form it has no fixed value, which is exactly why $\lim_{x \to 0^+} x^x$ takes work. The answer here is 1, but that is a fact about $x^x$ , not about the symbol $0^0$ . Other functions with the same form converge to other numbers, and some run off to $\infty$ .

Why is there no left-hand limit?

$x^x$ is really $e^{x \ln x}$ , and $\ln x$ needs $x > 0$ . Negative bases produce real values only at isolated exponents such as $-\frac{1}{3}$ , so there is no interval to the left of 0 on which the function exists.

Can I use L'Hopital directly on x^x?

No. The rule applies to quotients in $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form, and $x^x$ is neither. Taking the logarithm gives the product $x \ln x$ , and rewriting that as $\frac{\ln x}{\frac{1}{x}}$ makes the rule legal.