For what values of $r$, $x^r$ has infinite slope at $x=0$?

Question

I'm learning calculus form MIT OCW 18.01SC. In session 23 (it's about linear approximation), prof computes linear approximation near $0$ of some basic functions.
$$\sin{x}, \cos{x}, e^x, \ln{(1+x)}, (1+x)^r$$ Why two of them are shifted by $1$? I can see why we can't compute linear approximation of $\ln{x}$ near $0$, but why not $x^r$? Prof says "If you try to graph $x^r$, you'll discover that sometimes the slope is infinite, and so forth".

$\frac{d}{dx}x^r=rx^{r-1}$, how this is sometimes infinite (for not infinite $r$ and $x$)?

Answer 1

Think of the domain of this function for various values of r, and their corresponding graphs. For instance, r=-1.

http://www.sagemath.org/calctut/pix-calctut/onesided05.png

Answer 2

Consider negative values of r. Let r=-1. Then $d \over dx$$x^{-1}$= $-x^{-2}$=$-1 \over x^2$.

What happens when x=0?

Answer 3

As you wrote $$\frac{d}{dx}x^r=rx^{r-1}$$ there is no problem at $x=0$ if $(r-1) \geq 0$ that is to say if $r \geq 1$. But now, consider the case where $r \lt 1$. So the resulting exponent in the rhs becomes negative that is to say that the derivative looks like $\frac{1}{x^q}$ with $q \gt 0$ and you are looking at the slope at $x=0$.