Results giving Coninuity

Just as for sequences, building continuity directly by calculating limits soon becomes hard work. Instead we give a result that enables us to build new continuous functions from old ones, just as we did for sequences. Note that if f and g are functions and k is a constant, then k.f, f + g, fg and (often) f /g are also functions.

Proposition 4.16   Let f and g be continuous at a, and let k be a constant. Then k.f, f + g and fg are continuous at f. Also, if g(a)$\ne$ 0, then f /g is continuous at a.

Proof. We show that f + g is continuous at a. Since, by definition, we have (f + g)(a) = f (a) + g(a), it is enough to show that

$$\lim_{{x \to a}}^{}$$(f (x) + g(x)) = f (a) + g(a).

Pick $\epsilon$ > 0; then there is some $\delta_{1}^{}$ such that if | x - a| < $\delta_{1}^{}$, then | f (x) - f (a)| < $\epsilon$/2. Similarly there is some $\delta_{2}^{}$ such that if | x - a| < $\delta_{2}^{}$, then | g(x) - g(a)| < $\epsilon$/2. Let $\delta$ = min($\delta_{1}^{}$,$\delta_{2}^{}$), and pick x with | x - a| < $\delta$. Then

| f (x) + g(x) - (f (a) + g(a))|$$\le$$| f (x) - f (a)| + | g(x) - g(a)| < $$\epsilon$$/2 + $$\epsilon$$/2 = $$\epsilon$$.

This gives the result. The other results are similar, but rather harder; see (Spivak, 1967) for proofs. $\qedsymbol$

Note:Just as when dealing with sequences, we need to know that f /g is defined in some neighbourhood of a. This can be shown using a very similar proof to the corresponding result for sequences.

Proposition 4.17   Let f be continuous at a, and let g be continuous at f (a). Then gof is continuous at a

Proof. Pick $\epsilon$ > 0. We must find $\delta$ > 0 such that if | x - a| < $\delta$, then g(f (x)) - g(f (a))| < $\epsilon$. We find $\delta$ using the given properties of f and g. Since g is continuous at f (a), there is some $\delta_{1}^{}$ > 0 such that if | y - f (a)| < $\delta_{1}^{}$, then | g(y) - g(f (a))| < $\epsilon$. Now use the fact that f is continuous at a, so there is some $\delta$ > 0 such that if | x - a| < $\delta$, then | f (x) - f (a)| < $\delta_{1}^{}$. Combining these results gives the required inequality. $\qedsymbol$

Example 4.18   The function in Example 4.8 is continuous everywhere. When we first studied it, we showed it was continuous at the ``difficult'' point x = 0. Now we can deduce that it is continuous everywhere else.

Example 4.19   The function f : x $\longmapsto$ sin³x is continuous.

Solution. Write g(x) = sin(x) and h(x) = x³. Note that each of g and h are continuous, and that f = goh. Thus f is continuous.

Example 4.20   Let f (x) = tan$$\left(\vphantom{\frac{x^2-a^2}{x^2 + a^2}}\right.$$$${\frac{{x^2-a^2}}{{x^2 + a^2}}}$$$$\left.\vphantom{\frac{x^2-a^2}{x^2 + a^2}}\right)$$. Show that f is continuous at every point of its domain.

Solution. Let g(x) = $${\frac{{x^2-a^2}}{{x^2 + a^2}}}$$. Since -1 < g(x) < 1, the function is properly defined for all values of x (whilst tan x is undefined when x = (2k + 1)$\pi$/2 ), and the quotient is continuous, since each term is, and since x² + a²$\ne$ 0 for any x. Thus f is continuous, since f = tanog.

Exercise 4.21   Let f (x) = exp$$\left(\vphantom{\frac{1+x^2}{1-x^2}}\right.$$$${\frac{{1+x^2}}{{1-x^2}}}$$$$\left.\vphantom{\frac{1+x^2}{1-x^2}}\right)$$. Write down the domain of f, and show that f is continuous at every point of its domain.

As another example of the use of the definitions, we can give a proof of Proposition 2.20

Proposition 4.22   Let f be a continuous function at a, and let a_n$\to$a as n$\to$$\infty$. Then f (a_n)$\to$f (a) as n$\to$$\infty$.

Proof. Pick $\epsilon$ > 0 we must find N such that | a_n - f (a)| < $\epsilon$ whenever n$\ge$N. Now since f is continuous at a, we can find $\delta$ such that if | x - a| < $\delta$, then | f (x) - f (a)| < $\epsilon$. Also, since a_n$\to$a as n$\to$$\infty$ there is some N (taking $\delta$ for our epsilon -- but anything smaller, like $\delta$ = $\epsilon$/2 etc would work) such that | a_n - a| < $\delta$ whenever n$\ge$N. Combining these we see that if n$\ge$N then | a_n - f (a)| < $\epsilon$, as required. $\qedsymbol$

We can consider the example f (x) = sin(1/x) with this tool.

Example 4.23   Suppose that $\lim_{{x \to 0}}^{}$sin(1/x) = l; in other words, assume, to get a contradiction, that the limit exists. Let x_n = 1/($\pi$n); then x_n$\to$ 0 as n$\to$$\infty$, and so by assumption, sin(1/x_n) = sin(n$\pi$) = 0$\to$l as n$\to$$\infty$. Thus, just by looking at a single sequence, we see that the limit (if it exists) can only be l. But instead, consider the sequence x_n = 2/(4n + 1)$\pi$, so again x_n$\to$ 0 as n$\to$$\infty$. In this case, sin(1/x_n) = sin((4n + 1)$\pi$/2) = 1, and we must also have l = 1. Thus l does not exist.

Note:Sequences often provide a quick way of demonstrating that a function is not continuous, while, if f is well behaved on each sequence which converges to a, then in fact f is continuous at a. The proof is a little harder than the one we have just given, and is left until next year.

Example 4.24   We know from Prop 4.22 together with Example 4.8 that if a_n$\to$ 0 as n$\to$$\infty$, then

a_nsin$$\left(\vphantom{\frac{1}{a_n}}\right.$$$${\frac{{1}}{{a_n}}}$$$$\left.\vphantom{\frac{1}{a_n}}\right)$$$$\to$$0    as    n$$\to$$$$\infty$$.

Prove this directly using squeezing.

Solution. The proof is essentially the same as the proof of Example 4.8. We have

0$$\le$$$$\left\vert\vphantom{a_n \sin{\frac{1}{a_n}}}\right.$$a_nsin$${\frac{{1}}{{a_n}}}$$$$\left.\vphantom{a_n \sin{\frac{1}{a_n}}}\right\vert$$ = | a_n|.$$\left\vert\vphantom{ \sin{\frac{1}{a_n}}}\right.$$sin$${\frac{{1}}{{a_n}}}$$$$\left.\vphantom{ \sin{\frac{1}{a_n}}}\right\vert$$$$\le$$| a_n|$$\to$$0    as    n$$\to$$$$\infty$$.