# Quod Erat Demonstrandum

## 2009/03/18

### Continuity of composite functions

Filed under: HKALE,Pure Mathematics — johnmayhk @ 7:50 上午

It is well-known that the following is NOT true in general,

$\lim_{x \rightarrow a}f(g(x)) = f(\lim_{x \rightarrow a}g(x))$

Just give an example, let

$f(x) = \frac{x^2 - 1}{x - 1}$
$g(x) = \frac{\sin x}{x}$

Easy to have

$\lim_{x \rightarrow 0}f(g(x)) = 2$, but
$f(\lim_{x \rightarrow 0}g(x))$ is undefined, hence

$\lim_{x \rightarrow a}f(g(x)) \ne f(\lim_{x \rightarrow a}g(x))$ in this case.

So, when does the equality sign hold? Here is a simple theorem in secondary school pure mathematics.

Suppose
$g(x)$ is continuous at $x = a$ and
$f(x)$ is continuous at $x = g(a)$,
then
$f(g(x))$ is continuous at $x = a$.

It may be a bit confusing when using $x$ as the (independent) variable of both functions $f$ and $g$, though it is perfectly OK. For educational purpose, may be it is better to re-write the theorem as follows.

Suppose
$g(x)$ is continuous at $x = a$ and
$f(t)$ is continuous at $t = g(a)$,
then
$f(g(x))$ is continuous at $x = a$.

There is a ‘bridge’ $t$, which is also a fuction of $x$, that is $t = g(x)$.

Let’s prove the theorem step by step.

The given that “$g(x)$ is continuous at $x = a$" can be interpreted in this way.

When the value of $x$ tends to $a$, the value of $g(x)$ will get closer and closer to that of $g(a)$.

In symbol,

$x \rightarrow a \Rightarrow g(x) \rightarrow g(a)$ – – – – – – (*)

Besides, the given that “$f(t)$ is continuous at $t = g(a)$“, by the definition of continuity, we have

$\lim_{t \rightarrow g(a)}f(t) = f(g(a))$ – – – – – – (**)

OK, to show that “$f(g(x))$ is continuous at $x = a$“, we need to show

$\lim_{x \rightarrow a}f(g(x)) = f(g(a))$

Hence, we start with $\lim_{x \rightarrow a}f(g(x))$.

$\lim_{x \rightarrow a}f(g(x))$
$= \lim_{x \rightarrow a}f(t)$ [where $t = g(x)$]
$= \lim_{g(x) \rightarrow g(a)}f(t)$ [by (*)]
$= \lim_{t \rightarrow g(a)}f(t)$
$= f(g(a))$ [by (**)]

That is $f(g(x))$ is continuous at $x = a$.

Of course, it is extremely easy to give a proof using $\epsilon - \delta$ language. But, again, this is for secondary school educational purpose, right?

OK, the conclusion above is

$\lim_{x \rightarrow a}f(g(x)) = f(g(a))$ for $f, g$ satisfying certain conditions.

Knowing that $g(a) = \lim_{x \rightarrow a}g(x)$, the result can be written as

$\lim_{x \rightarrow a}f(g(x)) = f(\lim_{x \rightarrow a}g(x))$

and hence the original question is answered.

[SBA time]
1. Try to find out loopholes of the so-called proof above. Give comments on the weakness of the discussion above.

2. Try to visualize and explore the situation by clicking the following created by a silly teacher named John Ng.

[Enrichment]
1. Suppose $f, g$ are continuous at $x = a$, it is easy to know that $f+g, f-g, fg$ and $\frac{f}{g}$ ($g(a) \ne 0$) are continuous at $x = a$.

How about $f \bigvee g$ and $f \bigwedge g$? Are they continuous at $x = a$ as well?

Here
$f \bigvee g = \max\{f , g\}$
$f \bigwedge g = \min\{f , g\}$

(e.g. $f(x) = x + 3, g(x) = 2x$, then $(f \bigvee g)(4) = 8, (f \bigwedge g)(4) = 7$)

The answer is affirmative. And the proof is extremely simple when knowing that

$f \bigvee g$ and $f \bigwedge g$ can be converted into

$f \bigvee g = \frac{1}{2}(f + g + |f - g|)$
$f \bigwedge g = \frac{1}{2}(f + g - |f - g|)$

Try to verify the statements above.

2. Do you know that “continuous" can be refined into “upper semi-continuous" and “lower semi-continuous"? Try to explore the concepts by reading books and searching from the internet.

## 2 則迴響 »

1. I wonder if there should be no absolute signs on f + g for maximum and minimum functions…

迴響 由 皮旦 — 2009/03/19 @ 8:00 上午 | 回應

2. Yes, thank you 皮旦!

迴響 由 johnmayhk — 2009/03/19 @ 8:07 上午 | 回應