Limit of $x^x$ without L'Hôpital I was trying to calculate $\lim_{x \to 0^{+}} x^x$ without L'Hôpital's rule but could not make progress.
My best shot was to show that $\lim_{x \to 0^{+}} x\ln x = 0$ as that would imply the first limit.
Can anyone help me?
 A: Hint: What else do you know about $\ln x$? How does its growth rate compare to $\frac1x$ or $x$?
Another hint: you can transform this to a $\lim_{y \rightarrow \infty}$ problem i.e. by setting $y=1/x$. I find this much easier to work with. Thinking about things close to $0^+$ is hard.
You will want to use some combination of finding upper / lower bounds, and technically you will apply the squeeze theorem.
A: To do this from scratch, here's one approach:
First note that $x\ln x<0$ as soon as $x<1$.
Next, observe that $ne^{-n}\to 0$ as $n\to \infty ,\ $ because $ne^{-n}<\frac{n}{1+n+\frac{n^{2}}{2}}$.
And since $(x\ln x)'=\ln x+1,\ h(x)=x\ln x$ is decreasing on $(0,e^{-n})\ $ for all integers $n\geq 1$, which means we must have $x\ln x>-ne^{-n}$ for all $x\in (0,e^{-n})$.
Putting this all together then, we can write 
$-ne^{-n}<x\ln x<0$ whenever $x\in (0,e^{-n})$ and now to finish, we need only appeal to the squeeze theorem to obtain the result. 
A: In THIS ANSWER I showed using on the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities
$$\frac{x-1}{x}\log(x)\le x-1 \tag 1$$
for $x>0$.  
Using $\log(x^{\alpha})=\alpha \log(x)$ for all $\alpha$, we have for any $\alpha >0$ and $x>0$
$$\frac{x-x^{1-\alpha}}{\alpha}\le x\log(x)\le \frac{x^{1+\alpha}-x}{\alpha}$$
whereupon choosing $\alpha <1$ and applying the squeeze theorem reveals
$$\lim_{x\to 0^+}\left(x\log(x)\right)=0$$
Finally, we have
$$\begin{align}
\lim_{x\to 0^+}x^x&=\lim_{x\to 0^+}e^{x\log(x)}\\\\
&=e^{\lim_{x\to 0^+}\left(x\log(x)\right)}\\\\
&=1
\end{align}$$
TOOLS USED: 
(i)   The limit definition of the exponential function;
(ii)  Bernoulli's Inequality;
(iii) $\log(x^{\alpha})=\alpha \log(x)$;
(iv)  The Squeeze Theorem;
(v)   Continuity of the exponential function;
(vi)  $\lim_{x\to a}f(g(x))=f\left(\lim_{x\to a}g(x)\right)$ when $f$ is continuous.
A: Consider $A=x^x$. Taking logarithms $\log(A)=x\log(x)$ which goes to $0$ if $x\to 0$. So $\log(A)\to 0 \implies A \to 1$.
