Limit of $x^x$ as $x$ tends to $0$

Question

I am trying to solve the following limit:

$$\lim \limits_{x\to0} x^x$$

The only thing that comes to mind is to write $x^x$ as $e^{x\ln{x}}$ and getting the right sided limit would be easy but I don't see how I could get the left sided one seeing that the $\ln$ is not defined for negative numbers.

Is there something I am missing or is there another way to go about it?

P.S.:I don't know anything about derivatives so please keep it to the limits.

Answer 1

Claim: $\lim_{z \to 0} z^z = 1$, no matter which branch of the logarithm is used to define $z^z$.

More rigorously: let $U \subset \Bbb{C}$ be any simply connected open set such that $0 \in \overline{U} \setminus U$. Since $U$ is simply connected, there exists a branch of the natural logarithm $\ln : U \to \Bbb{C}$ defined on $U$. This branch corresponds uniquely to a branch of $f(z) = z^z$ on $U$ given by defining $z^z := e^{z \ln z}$. No matter what branch is selected, we will have $$\lim_{z \to 0} z \ln z = 0,$$ as we can see by switching to polar form. Every $z \in U$ may be written uniquely as $z = r e^{i \theta}$, where $r := r(z) = |z|$ is independent of the choice of branch of $\ln$ and $\theta := \theta(z) = \operatorname{Im}(\ln(z))$ is uniformly bounded on all of $U$. If we write $\ln(z) := \ln(r(z)) + i \theta(z)$, then as $z \to 0$, $r(z) \to 0$ and \begin{align*} \lim_{z \to 0} z \ln z &= \lim_{r \to 0^+} re^{i \theta} (\ln(r) + i \theta) \\ = \lim_{r \to 0^+} [(r \ln r) e^{i \theta}] &+ \lim_{r \to 0^+} (i r \theta e^{i \theta}) \\ = 0 &+ 0 = 0, \ \end{align*}

since $\lim_{r \to 0^+} (r \ln r) = 0$ and the functions of $\theta$ are bounded everywhere on $U$. It follows that $$\lim_{z \to 0} f(z) = \lim_{z \to 0} e^{z \ln z} = e^{\lim_{z \to 0} z \ln z} = e^0 = 1.$$

Answer 2

L'Hospital's rule is quickest. I show that other approaches are possible:

For $\ x\in \left[\ \frac{1}{7},\ \frac{1}{6}\ \right),$

$$ \left( \frac{1}{7} \right)^{\frac{1}{6}}<x^x<1.$$

Now using Newton's Binomial expansion,

\begin{align} \left(1-\alpha\right)^\frac{1}{6} = 1 + \frac{1}{6}(-\alpha) + \frac{\left(\frac{1}{6}\right)\left(-\frac{5}{6}\right)}{2!}(-\alpha)^2 + \frac{\left(\frac{1}{6}\right)\left(-\frac{5}{6}\right)\left(-\frac{11}{6}\right)}{3!}(-\alpha)^3+\ldots \\ \\ = 1 - \frac{1}{6}\alpha - \frac{\left(\frac{1}{6}\right)}{2}\frac{\left(\frac{5}{6}\right)}{1} \alpha^2 - \frac{\left(\frac{1}{6}\right)}{3} \frac{\left(\frac{11}{6}\right)}{2} \frac{\left(\frac{5}{6}\right)}{1} \alpha^3 - \ldots\ \\ \\ > 1 - \frac{1}{6}\left(\alpha + \frac{1}{2} \alpha^2 + \frac{1}{3} \alpha^3 + \ldots \right)\\ \\ = 1 - \frac{1}{6}(\ -\ln(1-\alpha)\ ) = 1 + \frac{1}{6}(\ \ln(1-\alpha)\ ).\\ \\ \end{align}

Substituting $\ \alpha = \frac{6}{7},\ $ into the above, we see that

$$ 1+ \frac{1}{6}\ln\left(\frac{1}{7}\right) = 1 - \frac{1}{6}\ln(7) < \left( \frac{1}{7} \right)^{\frac{1}{6}}<x^x<1\quad \forall x\in \left[\ \frac{1}{7},\ \frac{1}{6}\ \right).$$

More generally, for any $\ k \in\mathbb{N},\ $ we have,

$$ 1 - \frac{\ln(k+1)}{k} < \left( \frac{1}{k+1} \right)^{\frac{1}{k}} < x^x < 1 \quad \forall x\in \left[\ \frac{1}{k+1},\ \frac{1}{k}\ \right).$$

Letting $\ k\to\infty\ $ yields the result.