**Remark**: According to this definition, in order for to be continuous at
we
must have

and

The second condition is often not included in the definition of continuity, so this definition does not quite correspond to the usual definition.

**Remark:** The method we will usually use to show that
a function is *not* continuous at a point , is to find a
sequence in
such that
, but either diverges or converges
to a value different from .

proved in lemma 11.17 that a function is continuous at every point at which it is differentiable. (You should now check the proof of that lemma to see that we did prove this.) Hence , , , and (for ) are all continuous on their domains, and if , then is continuous on .

Our limit rules all give rise to theorems about continuous functions.

Proof: Suppose and are continuous at , and
is approachable from
. Then

By the sum rule for limits (theorem 10.15) it follows that

Thus is continuous at . The proofs of the other parts of the theorem are similar.

where is a subset of such that every subinterval of of positive length contains a point in and a point not in . Let .

- Case 1.
- If we can find a sequence of points in
such that . Then

so is not continuous at . - Case 2.
- If we can find a sequence of points in
such
that . Then

so is not continuous at .

I claim that is continuous. We know that is differentiable on , so is continuous at each point of . In example 10.13 we showed that so is also continuous at .

Then and are both continuous functions. Now

and hence is not continuous. The domain of contains just one point, and that point is not approachable from .

Proof: Suppose is continuous at and is continuous at , and
is
approachable from
. Let be a
sequence in
such
that . Then
since is continuous at .
Hence
since is continuous at ;
i.e.,

Hence is continuous at .