Continuity of Compositions

If functions $f(x),g(x)$ are continuous, what can we say about their composition $f\circ g = f(g(x))$?

Yes! Now let's see why. Suppose we want to study $f(g(x))$ for $x$ close to $p$ (e.g. how close is $f(g(x))$ to $f(g(p))$?).

Let's first think about the inner function $g$ at or around $p$ (since $g(p)$ is an important part of $f(g(p))$). We know $g$ is continuous at $p$, or in other words, that $g(p)$ is a good approximation for $g(x)$ when $x$ is close to $p$. So let's record that: $$g(x)\approx g(p)\text{ when } x\approx p.$$

Now we want to use the continuity of $f$. But continuity at what point?

OK, let's first write down what we know about $f$, and choose the point later. Continuity at a point $q$ means that $f(y)\approx f(q)$ when $y\approx q$ (we write $y,q$ now to avoid re-using variables). We want to end up with a statement like "$f(g(x))\approx f(g(p))$", so let's take $q = g(p)$: $$f(y) \approx f(g(p))\text{ when } y\approx g(p).$$

How can we synthesize what we've learned so far?

Let's revisit our two main results so far: $$\begin{align*} g(x) &\approx g(p) \text{ when } x\approx p; \\ f(y) \approx f(g(p)) \text{ when } y &\approx g(p). \end{align*}$$

Ah, but the solution's now staring us in the face!

Indeed, "substituting" $y = g(x)$, we find that $$f(g(x))\approx f(g(p)) \text{ when } g(x) \approx g(p), \text{ when } x \approx p.$$

Slightly more precisely,

$f(g(x))$ is close$_1$ to $f(g(p))$ if $g(x)$ is close$_2$ enough to $g(p)$, which is in turn guaranteed if $x$ is close$_3$ enough to $p$.

Here close$_2$ "depends" on close$_1$, and close$_3$ "depends" on close$_2$. Finally, a formal proof:

Fix $\epsilon_1>0$. Then $f(g(x))$ is at most $\epsilon_1$ away from $f(g(p))$ if $g(x)$ is at most some $\epsilon_2 > 0$ (depending on $\epsilon_1$) away from $g(p)$, which is in turn guaranteed if $x$ is at most some $\epsilon_3 >0$ (depending on $\epsilon_2$) away from $p$.

(Usually people use $\delta$ instead of $\epsilon_2$, etc., but that's no big deal.)

Of course, there's no reason to limit ourselves to a single composition. The same reasoning shows that something like $A(B(C(x)))$ is continuous at a point $p$ if $C(x)$ is continuous at $p$, $B(x)$ is continuous at $C(p)$, and $A(x)$ is continuous at $B(C(p))$.

Hope this made sense!