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\to8}f(x)# of the function \[f(x)=x^2+9x+6\]

#\displaystyle\lim_{x\to8}f(x)=142#

The function #f# is continuous and defined in the point #x=8#; hence \[\begin{aligned}\lim_{x\to8}f(x)&=f(8)\\[0.25cm]&=8^2+9\cdot 8+6\\[0.25cm]&=142\end{aligned}\]

New example