How do I calculate $\lim_{x\rightarrow 0} x\ln x$ I was thinking about the reasons behind $0^0=1$ and I remember one of my friends studying math arguing about the continuity of the function $x^x$ in $0$. But when I write as $$x^x=e^{x\ln x}$$ I am now looking at $$\lim_{x\rightarrow 0} x\ln x$$
Graphically I can see in Mathematica that it goes to $0.$ But I can't calculate by using a Taylor expansion, because I can't expand log around $0$. How do you prove that?
 A: L'Hopital's rule can be used,
$$\lim_{x\to 0^+}x\ln x=\lim_{x\to 0^+}\frac{\ln x}{\frac{1}{x}}$$
Using L'Hopital's Rule (as it has$\frac{-\infty}{\infty } $ form, and $\ln(x)$ and $1/x$ are differentiable to the right of $0$) we have,
$$\lim_{x\to 0^+}\frac{\ln x}{\frac{1}{x}}=\lim_{x\to 0^+}\frac{\frac{d\ln x}{dx}}{\frac{d\frac{1}{x}}{dx}}=\lim_{x\to 0^+}\frac{\frac{1}{x}}{\frac{-1}{x^2}}=\lim_{x\to 0^+}(-x)=0$$
A: Substitute $x = e^y$, then if $x \to 0$, $y\to -\infty$. So the limit becomes $$\lim\limits_{x\to 0} x \ln x = \lim\limits_{y\to -\infty} y\cdot e^{y}  = 0, $$ because $e^\cdot$ is stronger than any polynomial.
A: $ln(x) =  -\sum_{k=1}^\infty {(-1)^k (x-1)^k \over k}$  for $|x-1|<1$
Then $xln(x) =  -\sum_{k=1}^\infty {x(-1)^k (x-1)^k \over k}$ wich converges for $x=0$.
q.e.d.
A: Putting $y = 1/x$,
$x^x
=(1/y)^{1/y}
=\dfrac1{y^{1/y}}
$.
So,
if we can show that
$\lim_{y \to \infty} y^{1/y}
= 1$,
then $\lim_{x \to 0} x^x = 1$.
Here (again)
is a simple proof that
$n^{1/n} \to 1$
as $n \to \infty$.
By Bernoulli's inequality,
$(1+n^{-1/2})^n
\gt 1+n^{-1/2}n
=1+n^{1/2}
> n^{1/2}$.
Raising to the
$2/n$ power,
$(1+n^{-1/2})^2
> n^{1/n}$
so that
$n^{1/n}
< (1+n^{-1/2})^2
= 1+2n^{-1/2} + n^{-1}
< 1+3n^{-1/2}
$.
Since
$n^{1/n} > 1$,
$n^{1/n} \to 1$.
A: $(x\ln x)'=\ln x+1\to -\infty$ when $x\to 0^+$ we conclude that $(x\ln x)$ is a decreasing function near $0$.
Since $x\ln(x)\le 0$ then for $x$ sufficiently small we have $|x\ln x|<C$ for some constant $C$.
Let's have $x\to 0^+$, $y\to 0^+$ so that $u=xy\to 0^+$.
$|u\ln u|=|xy\ln(xy)|=|xy(\ln x+\ln y)|=|y(x\ln x)+x(y\ln y)|\le|y(x\ln x)|+|x(y\ln y)|\le C(|x|+|y|)\to 0$
