# Rule de l'Hôpital - Continuous and Differentiable

## Homework Statement

Prove that

##h: [0,\infty) \rightarrow \mathbb R, x \mapsto
\begin{cases}
x^x, \ \ x>0\\
1, \ \ x = 0\\
\end{cases}
##
is continuous but not differentiable at x = 0.

## The Attempt at a Solution

To show continuity, the limit as x approaches 0 from the right must equal to 1. Meaning:

##\lim_{x \ \downarrow \ 0} h(x) = 1##.

##\lim_{x \ \downarrow 0} h(x) = \lim_{x \ \downarrow 0} x^x = 0^0 = 1## (I'm not too content with this part. Just plugging in 0 seems wrong to me and I would have to maybe use the definition of continuity?)

Differentiable:
##h'(0) = 0##
##\lim_{x \ \downarrow 0} \frac{f(x) - f(0)}{x-0} = \lim_{x \ \downarrow 0} \frac{x^x - 0^0}{x-0} = \lim_{x \ \downarrow 0} \frac{x^x - 1}{x} = \lim_{x \ \downarrow 0} x^{x-1}-1## does not exists due to 1/0. Therefore h(x) is not differentiable.

Last edited:

Orodruin
Staff Emeritus
Homework Helper
Gold Member
There is no left side. The function is not defined for ##x < 0##.

Crap. Right. Fixed it.

Can anyone give me some insight please?

I made an error in my calculations. I think I got it now.

Dick