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.
First find $\lim\limits_{x\rightarrow 0}x\ln(x)=\lim\limits_{x\rightarrow 0}\frac{\ln(x)}{1/x}$. Using L'Hospital this become $\lim\limits_{x\rightarrow 0}\frac{1/x}{-1/x^{2}}=\lim\limits_{x\rightarrow 0}-x=0$.
So $\lim\limits_{x\rightarrow 0}e^{x\ln(x)}=e^{\lim\limits_{x\rightarrow 0}x\ln(x)}=1$.
(note assuming $x>0$ of course, since $x^{x}$ is not well-defined otherwise)
Also, if you allow $x<0$ but $x$ must be rational only, then the limit do not exist. This can be seen from the fact that $\lim\limits_{x\rightarrow 0}x^{x}=1$ when $x>0$. This means, that there are positive $x$ arbitrarily close to $1$ in any neighbourhood of $0$. Now once you look at negative $x$, then by considering rationals with odd denominator in any neighbourhood of $0$, then the once with even numerator will be positive, and odd numerator will be negative. The positive one will be arbitrary close to $1$, the negative one will be arbitrarily close to $-1$, so there are no limit.
If you allow $x<0$ and $x$ must be rational only, but also allow only a subset of rational such that $x^{x}$ have definite sign, then the limit is either $1$ or $-1$ from the left.
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 $$\lim_{z \to 0} f(z) = \lim_{z \to 0} e^{z \ln z} = e^{\lim_{z \to 0} z \ln z} = e^0 = 1.$$
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.
