Let's suppose that we have two real functions of a real variable, `f(x)`
and `g(x)`. Since they are both real functions of a real variable,
we can form the composite function, `f(g(x))`, which is simply the
operation of applying `g` to `x`, then applying `f`
to the result.

Let's also suppose that `g(x)` is continuous at a particular point,
` x = a``f(x)` is continuous at
the point,
` x = g(a)`

We shall prove that `f(g(x))` is continuous at
` x = a`

By the limits definition of continuity, we know that:

lim g(x) = g(a) eq. 3.1a-1a xand~~> a~~

lim f(x) = f(g(a)) eq. 3.1a-1b xWhat we would like to prove is:~~> g(a)~~

lim f(g(x)) = f(g(a)) eq. 3.1a-2 xIf you don't understand how I came up with 3.1a-1a through 3.1a-2, then you should review section 3.0 carefully, paying special attention to what the limit definition of continuity means. These three equations can be derived simply by symbol substitution from 3.0-1 or 3.0-2. 3.1a-1a is straightforward. In 3.1a-1b, we are saying that as~~> a~~

As always, we return to our contract interpretation of limits. The contract
for 3.1a-1a is: for any
` e _{g} > 0 `

|g(x) - g(a)| £ ewhenever_{g}eq. 3.1a-3a

Likewise, the contract for 3.1a-1b is: for any
` e _{f} > 0 `

|f(x) - f(g(a))| £ ewhenever_{f}eq. 3.1a-3b

Now suppose that you name an
`epsilon _{f} > 0`

|f(g(x)) - f(g(a))| £ ewhenever_{f}eq. 3.1a-4

Well, 3.1a-3b says I already have a recipe to go from
`e _{f}` to
a

If you are still unsteady on `delta-epsilon` proofs, keep at it.
If you can't seem to get this one, then go back to some others that we
did previously and review. Then tackle this one again. Also try reading
the proofs out loud. That often helps. And remember, the likelihood
of a `delta-epsilon` proof turning up on an exam is high.

email me at *hahn@netsrq.com*