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I'm looking for a function $f$ that is continuous in $x$, i.e. for which holds $$\lim_{y\to 0} f(x+y)-f(x) = 0$$ , but for which its derivative
$$ \lim_{y\to 0} \frac {f(x+y)-f(x)}y = \infty$$ tends to infinity.

The whole idea of being continuous in a point, yet having an infinite gradient in this point seems like a contradiction, but as differentiable$\implies$continuous is only an implication, it should be possible.

As a possible candidate I've e.g. looked for solutions for the functional equation $f(x+y) = f(x) + g(y)$, where $f$ is continuous,
but as $f(x) = f(0+x) = g(x) $, the only solution for $f(x)$ is $c\cdot x$, which doesn't fulfill the requirement.

Sudix
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    For example $f(x)=\sqrt{x}$ at $x=0$. You may take also $x^a$ with $0<a<1$. – Robert Z Jun 26 '18 at 06:02
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    Probably drawing a picture will give you a better approach. There is a well-known elementary function with this property. – hardmath Jun 26 '18 at 06:02
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    https://math.stackexchange.com/questions/65770/infinite-gradient-and-continuity Might be relevant – asdf Jun 26 '18 at 06:04
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    Take $,f(x)=\sqrt[3]{x},$ for example. @RobertZ Your $f(x)$ is not defined on $\mathbb{R}^-$ which can introduce unnecessary confusion, depending on the exact definitions. – dxiv Jun 26 '18 at 06:08
  • Defined on R. f(x) = √x for x \ge 0 , f(x) = -√|x| for x \le 0. – Peter Szilas Jun 26 '18 at 06:14
  • Thanks guys! I kinda missed the forest for the trees it seems – Sudix Jun 26 '18 at 06:16

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Classic example (not the simplest possible, of course):

$$f(x)=\sqrt[3]{x^3-3x^2}$$

enter image description here

Infinite derivatives at $x=0$ and $x=3$, continuous everywhere. I still remember this function from my math exam, like 3 decades ago :)

Saša
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