Continuity of functions

Limits part 2: Functions: All kinds of limits

Continuity of functions

There is an important link between the different limits at a point.

For each $a\in\mathbb R$ and every $L\in \mathbb R \cup\{\pm\infty\}$ holds

$\lim_{x\to a} f(x) = L$ if and only if

$\lim_{x\uparrow a} f(x) = \lim_{x\downarrow a} f(x) = L$

For the math enthusiast, we prove the theorem only for finite $L$ .

" $\implies$ ": Suppose $\lim_{x\to a} f(x) = L$ . This means that for every number $\epsilon>0$ there exists a $\delta>0$ such that if $0<|x-a|<\delta$ then $|f(x) - L | <\epsilon$ . Let $\epsilon > 0$ be given and select a $\delta$ with the mentioned property. Then for all $x\in (a-\delta,a)$ holds $|f(x) - L | <\epsilon$ , because $(a-\delta,a)\subset (a-\delta,a+\delta)$ . This shows that $\lim_{x\uparrow a} f(x) = L$ . In exactly the same way holds for all $x\in (a, a+ \delta)$ that $|f(x) - L | <\epsilon$ . This proves that $\lim_{x\downarrow a} f(x)$ is also equal to $L$ , that is,

$\lim_{x\uparrow a} f(x) = \lim_{x\downarrow a} f(x) = L$

" $\impliedby$ ": Suppose that the other statement is true, that is,

$\lim_{x\uparrow a} f(x) = \lim_{x\downarrow a} f(x) = L$ Let $\epsilon > 0$ be given. According to the definition of $\lim_{x\uparrow a} f(x) = L$ there exists a number $\delta_1$ so that if $x\in (a - \delta_1, a)$ then $|f(x) - L|<\epsilon$ . By the definition of $\lim_{x\downarrow a} f(x) = L$ there is also a number $\delta_2$ such that if $x\in (a, a+\delta_2)$ then $|f(x) - L|<\epsilon$ . Now choose $\delta = \min\{\delta_1, \delta_2\}$ . For $x\in (a-\delta, a+\delta)$ unequal to $a$ then holds that $x\in (a-\delta,a)$ or $x\in (a,a+\delta)$ . In the first case we have $x\in (a-\delta_1,a)$ (because $\delta \leq \delta_1$ ) so that indeed $|f(x) - L | <\epsilon$ . In the second case we have $x\in (a,a+\delta)\subseteq (a,a+\delta_2)$ , so that also $|f(x) - L | <\epsilon$ . We conclude that in both cases $|f(x) - L | <\epsilon$ is true, and because $\epsilon>0$ was randomly chosen this implies

$\lim_{x\to a} f(x) = L$

Recall that the limit $\lim_{x\to a}f(x) = L$ says nothing about the value $f(a)$ because we only study $x$ values that meet the condition $0<|x-a|<\delta$ . When the function value in $a$ equals the limit at $a$ then we say that the function is continuous at $x=a$ .

A function $f$ is continuous at a point $x=a$ when $a$ is an interior point of the domain of $f$ and

$\lim_{x\to a} f(x) = f(a)$

Applying the theorem we just proved, this means that the left-hand limit and the right-hand limit both equal $f(a)$ .

We speak of a continuous function when the function is continuous at all points of its domain.

Elementary functions such as $x^n, \sin(x), \cos(x),$ and $e^x$ are continuous. Well-known functions such as $\ln (x)$ and $\tan(x)$ are not for each value of $x$ defined, but they are continuous wherever they are well-defined.

If $f$ and $g$ are both continuous at $x=a$ then $f+g$ is also continuous at $x=a$ .

If $f$ and $g$ both continuous at $x=a$ we have

$\lim_{x\to a} f(x) = f(a) \quad\text{ and }\quad \lim_{x\to a} g(x) = g(a)$ From the sum rule for limits now follows that

$\lim_{x\to a} \left(f(x) + g(x)\right) = f(a) + g(a)\text,$ that is $\lim_{x\to a} (f+g)(x) = (f+g)(a)$ .

It directly follows that the sum of continuous functions is also continuous. The composition of two continuous functions, in other words $(f\circ g)(x) := f(g(x))$ is also continuously when both functions are continuous.

If $g$ is continuous at $x=a$ and $f$ is continuous at $x=g(a)$ then $f\circ g$ is also continuous at $x=a$ .

From this it follows that the composition of continuous functions is again continuous. This means in essence that you can swap limits and continuous functions:

$\lim_{x\to a} f( g(x)) = f\left( \lim_{x\to a} g(x) \right)$

We close this paragraph with a few examples.

$\lim_{x\to\infty} \cos\left(\frac{1}{x}\right) = \cos (0) = 1$

If you already know that a function is continuous, then you can calculate limits in a simple manner. The example below illustrates this.

Calculate the limit $\displaystyle\lim_{x\to4}f(x)$ of the function

$f(x)=x^2+7x+2$

$\displaystyle\lim_{x\to4}f(x)=46$

The function $f$ is continuous and defined in the point $x=4$ ; hence

$\begin{aligned}\lim_{x\to4}f(x)&=f(4)\\[0.25cm]&=4^2+7\cdot 4+2\\[0.25cm]&=46\end{aligned}$

New example