How to calculate $\lim\limits_{x\to 0}\ x^x$? How do we calculate:
$$\lim_{x \to 0} x^x$$
 A: Rewrite $x^x=e^{x\ln x}$. Now you can approach the limit in the exponent as you would any other limit. Even l'Hospital works just fine if you write it as $\ln x/(1/x)$.
A: By definition of exponentiation and basic properties of logarithms, $$x^x = \exp(\log(x^x)) = \exp(x\log x ).$$
From continuity of $\exp$ we have
$$\lim_{x\to0^+}\exp(x\log x ) = \exp\left(\lim_{x\to 0^+}x\log x\right). $$
By a nice argument from @user17662 here, we have that 
$$\lim_{x\to0^+}x\log x=0. $$
Hence
$$\lim_{x\to0^+}x^x=\exp(0)=1.$$
A: Write: $x^x = e^{x\ln(x)}$
Next, compute the limit of $x\ln(x)$ using L'hopital's.
A: A different approach to this question is as follows.  
First, let's consider $\lim_{n\to\infty}n^{1/n}$. Since $(1+x)^n\ge\dfrac{n(n-1)}{2}x^2>\dfrac{n^2}{4}x^2$, for $n>2$, let $1+x=n^{1/n}$, we have $n\ge\dfrac{n^2}{4}(n^{1/n}-1)^2$. Hence, $0<n^{1/n}-1<\dfrac{2}{\sqrt{n}}$ . We conclude $\lim_{n\to\infty}n^{1/n}=1$.  
Therefore, $\lim_{x\to0+}x^x=\lim_{y\to+\infty}\dfrac{1}{y^{1/y}}$ can be estimate by the following inequality:$$[y]^{1/([y]+1)}<y^{1/y}<([y]+1)^{1/[y]}.$$ 
Show that $\lim_{y\to+\infty}y^{1/y}=1$.
A: Since the function $f(x)=x^x$ is uniformly continuous in a right neighbourhood of zero, it is enough to replace $x$ with $\frac{1}{n}$ and compute the limit:
$$ \lim_{n\to +\infty}\left(\frac{1}{n}\right)^{\frac{1}{n}}=\lim_{n\to +\infty}\frac{1}{n^{1/n}}=1.$$
For instance, the previous line follows from AM-GM:
$$ 1\leq n^{1/n} = \left(\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)\right)^{1/n} \leq\frac{n+H_{n-1}}{n}\leq 1+\frac{\gamma+\log n}{n}.$$
A: Hint:
Write $x^x=e^{xln(x)}$ and use the continuity of $e^x$ and l'hospitals rule.
A: $$ \lim_{x \to 0^{+}} x^{x} =  e^{\lim_{x \to 0^{+}} x \ln x} = e^{\lim_{x \to 0^{+}} \frac{\ln x}{1/x}} $$
Now, applying L'Hospital's Rule,
$$ = e^{\lim_{x \to 0^{+}} \frac{1/x}{-1/x^{2}} } = e^{\lim_{x \to 0^{+}} e^{-x}} = e^{0} = 1.$$
