Continuity of Functions with Vertical Tangents I'm running into some confusion regarding properties of continuous functions. I'm comfortable with the epsilon-delta definition of limits and the basic definition of continuity at a point $a$ ($\lim_{x\rightarrow a}f(x)$ must exist, $f(a)$ must exist, and the two must equal one another), but I'm frequently encountering the statement that a function is continuous if and only if "a small change in $x$ produces a small change in $f(x)$". This seems reasonable enough, but continuous functions with vertical tangents seem to present a contradiction to this assertion. For example, the function $f(x)=\sqrt[3]{x}$ is continuous at $x=0$ by the definition of continuity, but its derivative at $x=0$ is undefined because the line tangent to the curve $y=f(x)$ is vertical. Shouldn't this mean, then, that at $x=0$ a small change in $x$ produces an infinite change in $y$? I'm not sure if $\left.\frac{dy}{dx}\right|_{x=0}=\infty$ constitutes an abuse of notation, but certainly it is true that $f^\prime(0)$ is undefined and $$\lim_{x\rightarrow 0}f^\prime(x)=\infty\text{.}$$ So I'm not sure how this "small change in $x$/small change in $f(x)$" description of continuity holds in this particular case.  Can anyone help resolve this confusion for me?
 A: This is exactly why we use algebraic proofs, not words and pictures, to establish what is fact in mathematics. Here is a proper proof that $x\mapsto x^{1/3}$ is continuous at $0$.

Definition: Continuity
A function $f:D\subseteq \Bbb{R}\to\Bbb{R}$ is continuous at a point $x_0$ if $\forall \epsilon>0$, $\exists \delta>0$ such that $|x-x_0|<\delta\implies |f(x)-f(x_0)|<\epsilon$

In our example, let's test if this applies to $f:x\mapsto x^{1/3}$ at the point $x=0$.
$$|f(x)-f(x_0)|<\epsilon\text{ in our case,} ~~|x^{1/3}|<\epsilon$$
We'll check left and right continuity separately. For right continuity, we have
$$x^{1/3}<\epsilon\implies x=|x-0|<\epsilon^3$$
So we have found our delta - it is precisely $\epsilon^3$. For left continuity,
$$-x^{1/3}<\epsilon\implies x^{1/3}>-\epsilon\implies x>-\epsilon^3$$
I.e, as long as $x$ is within $\epsilon^3$ of zero the inequality holds.

To sum up in words, the phrase "a small change in $x$ produces a small change in $f(x)$" is kind of correct, but needs to be interpreted correctly. Your remark

but its derivative at $x=0$ is undefined because the line tangent to the curve $y=f(x)$ is vertical. Shouldn't this mean, then, that at $x=0$ a small change in $x$ produces an infinite change in $y$?

Is not correct. Yes, it is true that the derivative of $x^{1/3}$ at zero is undefined-
$$\lim_{x\to 0}\frac{x^{1/3}}{x}=\infty$$
But for any nonzero value, say $t$, the change is definitely finite.
$$\frac{t^{1/3}}{t}=t^{-2/3}$$
Which is well defined $\forall t\neq 0$.
