# Function continuity

$f:\mathbb{R} \to \mathbb{R}$ is continuous at $x_0$ if $\forall \varepsilon > 0 \ \exists \delta > 0$ such that $$\left|x - x_0\right| < \delta \Rightarrow |f(x) - f(x_0)| < \varepsilon \ \forall x.$$

• For every error $\epsilon$ I give you in $y$, you can give me a tolerance $\delta$ in $x$ to satisfy my error.
– user61527
Feb 6, 2014 at 16:12

One way to think about it is with the following visual. As the function input $x$ changes, it forces the function output $f(x)$ to change accordingly, and these changes can be fairly unpredictable for arbitrary functions $f:\Bbb R\to\Bbb R.$ Continuity of $f$ at $x_0$ means we can keep the output changes as small as we like--that is, make $|f(x)-f(x_0)|$ less than any positive $\epsilon$ that we like--just by making sure that the input changes are small enough--that is, by keeping $|x-x_0|<\delta$ for some small enough positive $\delta.$

Another way to visualize it is to think of $\epsilon$ as the length of a chain attached to $f(x)$ and $f(x_0)$; $δ,$ attached to $x$ and $x_0.$ For continuity, we need to know that no matter how short the $\epsilon$-chain is, we will always be able to make the $δ$-chain short enough so that $f(x)$ doesn't break the $\epsilon$-chain as $x$ runs around on the $\delta$-chain.

We won't always be able to find such a $\delta$ for all $\epsilon$ (though we may be able to find one for some $\epsilon$). For example, consider the function $f:\Bbb R\to\Bbb R$ given by $$f(x)=\begin{cases}0 & x=0\\1 & x\ne 0.\end{cases}$$ Let $x_0=0.$ Now, it's clear that no matter how far $x$ runs from $x_0,$ we will always be able to keep $f(x)$ within $2$ (for example) of $x_0.$ In fact, for any $\epsilon>1,$ we can take any $\delta>0,$ and be sure that $f(x)$ will always be within $\epsilon$ of $f(x_0)$ if $x$ is within $\delta$ of $x_0$. However, for $0<\epsilon\le 1,$ we can no longer do this. For example, take $\epsilon=\frac13.$ Now, taking any $\delta>0,$ we can let $x=\frac\delta2.$ Then $x=\frac\delta2$ is within $\delta$ of $x_0=0,$ but $f(x)=1$ is not within $\epsilon=\frac13$ of $f(x_0)=0.$

So, $f$ is not continuous at $x_0=0,$ even though some $\epsilon>0$ allow us to find an appropriate $\delta>0.$

It means that you can get $f(x)$ as close to $f(x_0)$ as you want by bringing $x$ close enough to $x_0$.

One way to think of continuous functions is that they are the functions that commute with limits: $$\lim\limits_{x\to x_0}f\left(x\right)=f\left(\lim\limits_{x\to x_0}x\right)$$

Probably it should help to draw this out. First thing you might try is drawing the graph of a function that does not satisfy this statement, i.e. is not continuous. Basically you will have to put a gap/jump somewhere in it.

The statement is then seen to mean simply that if you look at points on the x-axis coming closer to $X_0$ then the points on the graph should also come closer together, i.e. Xavier's argument about limits.