Limit of derivative does not exist, while limit of difference quotient is infinite Can anyone show an example of a function $f$ of a real variabile such that


*

*$f$ is differentiable on a neighborhood of a point $x_0 \in \mathbb{R}$, except at $x_0$ itself;

*$f$ is continuous at $x_0$;

*$\displaystyle \lim_{x \to x_0} f'(x)$ does not exist;

*$\displaystyle \lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0} = +\infty$ or $\displaystyle \lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0} = -\infty$ ?

 A: What about 
$$f(x)=\sqrt{x}+x\sin\frac{1}{x}$$
defined on $(0,+\infty)$, that can be defined by continuity at $0$ with $f(0)=0$? (I let you check that it works, if I am not wrong)
If you don't like it because it is only $\mathbb{R}_+$, you can replace with
$$f(x)=\sqrt[3]{x}+x\sin\frac{1}{x},$$
graph
that can be defined on $\mathbb{R}$.
Using google, you get a fairly obvious drawing explaining what is happening (personally before having an explicit example, I had a fairly good drawing in my head to convince myself that such an example was possible, and then help me to produce a formula)
A: How about $f(x)=\sqrt[3]{x}=x^{1/3}$ [Graph]?
$f'(x)=\dfrac{1}{3}x^{-2/3}$. Notice that $f'(x)$ exists everywhere except at $x=0$. Also, $f(x)=\sqrt[3]{x}$ is continuous everywhere.
However, $\lim\limits_{x \to 0} \dfrac{x^{1/3}-0^{1/3}}{x-0} = \lim\limits_{x \to 0} \dfrac{1}{x^{2/3}} = +\infty$.
While this curve is continuous, it has a vertical tangent at $x=0$ (resulting in the "infinite" derivative there).
[If you choose $f(x)=\sqrt[3]{x^2}=x^{2/3}$, you'll get a similar example where the limit definition of the derivative gives $\pm \infty$ as you approach $0$ from the left and right.]  
A: Imagine two semicircles joining when $x=x_0$ such as 
$$f(x) = \sqrt{1-(x_0+1-x)^2} \text{ when } x_0 \le x \le x_0+2$$
$$f(x) = -\sqrt{1-(x_0-1-x)^2} \text{ when } x_0-2 \le x \le x_0$$
Change the signs if you want alternative limits for the derivatives as $x \to x_0$
