108 CHAPTER 4. CONTINUITY AND LIMITS

The sine function is continuous. To see this, suppose limk→∞ xk = x

|sin(xk)− sin(x)|= |sin(x+ xk − x)− sin(x)|

= |sin(x)cos(xk − x)+ sin(xk − x)cos(x)− sin(x)|≤ |sin(x)| |cos(xk − x)−1|+ |cosx| |sin(xk − x)|≤ |cos(xk − x)−1|+ |sin(xk − x)|

Now, from the geometric definition of the sine and cosine, both terms on the right convergeto 0 as k → ∞. Thus the sine is continuous. Similar reasoning shows that the cosine iscontinuous.

Of course one can also consider the arcsin function defined as arcsin(y) is the anglewhose sine is y which is in

[−π

2 ,π

2

]. This is a well defined function because the sine

function is one to one on[−π

2 ,π

2

]. By Corollary 4.5.1 this inverse function is continuous.

The arccos function is defined on [−1,1] and arccos(y) is defined as the angle in [0,π]whose cosine is y. By the same corollary, this function is also continuous.

It was observed that a small change in x led to a small change in ln(x). Thus ln iscontinuous. It follows from Corollary 4.5.1 that its inverse exp is continuous. Then fromthe theorem about various combinations of continuous functions, Theorem 4.0.6, all of thefunctions logb for b ̸= 1, and all functions x → bx for b > 0 are continuous.

As noted above, all polynomials are continuous as are all rational functions at all pointsof their domain. Indeed, if p(x)/q(x) is a rational function and q(x0) ̸= 0, then if xn → x0,it follows that q(xn)→ q(x0) ̸= 0 and p(xn)→ p(x0). Then from the theorem on limits ofsequences, Theorem 3.3.7, it follows p(xn)/q(xn)→ p(x0)/q(x0).

It is now clear that we have a very large collection of functions which are known to becontinuous. The next chapter will consider something even better.

4.9 Sequences of FunctionsSuppose for each n ∈ N, fn is a continuous function defined on some interval [a,b] . Alsosuppose that for each fixed x ∈ [a,b] , limn→∞ fn (x) = f (x). This is called pointwise con-vergence. Does it follow that f is continuous on [a,b]? The answer is NO. Consider thefollowing

fn (x)≡ xn for x ∈ [0,1]

Then limn→∞ fn (x) exists for each x ∈ [0,1] and equals

f (x)≡

{1 if x = 1

0 if x ̸= 1

You should verify this is the case. This limit function is not continuous. Indeed, it has ajump at x = 1. Here are graphs of the first few of these functions.

0 0.5 1

x

0

0.5

1

y