Continuity of a Function

Continuity at a Point

Definition: A function $f$ is said to be Continuous at the value $a$ if $\displaystyle{\lim_{x \to a} f(x) = f(a)}$ which implies that:
a) $f(a)$ is defined.
b) $\lim_{x \to a} f(x)$ exists.
c) $\lim_{x \to a} f(x) = f(a)$.
If at least one of these conditions is not satisfied, then $f$ is said to be Discontinuous and has a Discontinuity at $a$.

For example, consider the following function $f(x) = \frac{1}{x}$. We know that there exists a discontinuity in this function at $x = 0$ since our function evaluated at our suspected value of discontinuity $f(0) = \frac{1}{0}$ is undefined.

Furthermore, $f$ does not satisfy the second condition for continuity since $\lim_{x \to 0^-} f(x) = -\infty$ and $\lim_{x \to 0^+} f(x) = \infty$. Therefore, $\lim_{x \to 0} f(x) \: \mathrm{D.N.E.}$.

Example 1

Consider the following piecewise function $f(x) = \left\{\begin{matrix} x^2 - \frac{2}{x} & \mathrm{if} \: x > -2\\ 4 & \mathrm{if} \: x \leq -2 \end{matrix}\right.$. At what points is $f$ discontinuous?

The following is a graph of $f$:

Screen%20Shot%202014-08-29%20at%206.29.21%20PM.png

We first suspect a discontinuity at $x = -2$. We note that $\lim_{x \to -2^-} f(x) = 4$, while $\lim_{x \to -2^+} f(x) = 3$. Therefore, a discontinuity exists at $x = -2$ by our definition.

We also suspect there to be a discontinuity at $x = 0$. We note that $f(0) = 0^2 + \frac{2}{0}$ is undefined. Since $f(0)$ is not defined, there exists a discontinuity at $x = 0$ as well.

Continuity over a Function's Domain and an Interval

Definition: A function $f$ is said to be Continuous if $f$ is continuous at every value $a \in D(f)$ (every value $a$ in the domain of $f$). Additionally, a function $f$ is continuous over an interval $I$ if $f$ is continuous at every value $a \in I$.

There are certain functions that are always continuous at every number in their domains, namely polynomials, trigonometric functions, exponential functions, logarithmic functions, root functions, etc…

Continuity Laws

We will now look at some very important laws of continuity.

Theorem 1: Let $f$ and $g$ be continuous functions at $a$, and $k$ is a constant. Then the following properties hold:
a). The sum $f + g$ is continuous at $a$.
b). The difference $f - g$ is continuous at $a$.
c). The multiple $kf$ is continuous at $a$.
d). The product $fg$ is continuous at $a$.
e). The quotient $\frac{f}{g}$ is continuous at $a$ provided that $g(a) \neq 0$.

We will now prove all five properties using the limit laws we have already looked at.

  • Proof of a): Since $f$ and $g$ are continuous at $a$, it follows that $\lim_{x \to a} f(x) = f(a)$ and $\lim_{x \to a} g(x) = g(a)$. Thus:
(1)
\begin{align} \lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) \\ \lim_{x \to a} [f(x) + g(x)] = f(a) + g(a) \\ \lim_{x \to a} [f(x) + g(x)] = (f + g)(a) \end{align}
  • Therefore, $f + g$ is continuous at $a$. $\blacksquare$
  • Proof of b): Since $f$ and $g$ are continuous at $a$, it follows that $\lim_{x \to a} f(x) = f(a)$ and $\lim_{x \to a} g(x) = g(a)$. Thus:
(2)
\begin{align} \lim_{x \to a} [f(x) - g(x)] = \lim_{x \to a} f(x) - \lim_{x \to a} g(x) \\ \lim_{x \to a} [f(x) - g(x)] = f(a) - g(a) \\ \lim_{x \to a} [f(x) - g(x)] = (f - g)(a) \end{align}
  • Therefore, $f - g$ is continuous at $a$. $\blacksquare$
  • Proof of c): Since $f$ is continuous at $a$, then $\lim_{x \to a} f(x) = f(a)$ and thus:
(3)
\begin{align} \lim_{x \to a} f(x) = f(a) \\ k \lim_{x \to a} f(x) = kf(a) \\ \lim_{x \to a} kf(x) = kf(a) \end{align}
  • Therefore, $kf$ is continuous at $a$. $\blacksquare$
  • Proof of d): Since $f$ and $g$ are continuous at $a$, it follows that $\lim_{x \to a} f(x) = f(a)$ and $\lim_{x \to a} g(x) = g(a)$. Thus:
(4)
\begin{align} \lim_{x \to a} [f(x)g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x) \\ \lim_{x \to a} [f(x)g(x)] = f(a) \cdot g(a) \\ \lim_{x \to a} [f(x)g(x)] = (f \cdot g)(a) \\ \end{align}
  • Therefore, $fg$ is continuous at $a$. $\blacksquare$
  • Proof of e): Since $f$ and $g$ are continuous at $a$, it follows that $\lim_{x \to a} f(x) = f(a)$ and $\lim_{x \to a} g(x) = g(a)$. Thus:
(5)
\begin{align} \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} \\ \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{f(a)}{g(a)} \\ \lim_{x \to a} \frac{f(x)}{g(x)} = (f/g)(a) \end{align}
  • Therefore, $\frac{f}{g}$ is continuous at $a$. $\blacksquare$

Continuous Composite Functions

Theorem 2: Suppose that $f$ is continuous at the value $b$ and $\lim_{x \to a} g(x) = b$. It thus follows that $\lim_{x \to a} f(g(x)) = f(b)$ or rather, $\lim_{x \to a} f(g(x)) = f(\lim_{x \to a} f(x))$.

This theorem should make sense. We note that as $x \to a$, $g(x) \to b$ since $\lim_{x \to a} g(x) = b$. Therefore, as $x \to a$, $f(g(x)) \to f(b)$ and thus $\lim_{x \to a} f(g(x)) = f(\lim_{x \to a} f(x))$.

Theorem 3: Suppose that $f$ is continuous at $g(a)$, and $g$ is continuous at $a$. Then the composition $f \circ g$ is continuous at $a$.
  • Proof: We know that $g$ is continuous at $a$ and therefore $\lim_{x \to a} g(x) = g(a)$. We also know that $f$ is continuous at $g(a)$. Applying theorem 1, we get that:
(6)
\begin{align} \lim_{x \to a} f(g(x)) = f(g(a)) \end{align}
  • Therefore, $f \circ g = f(g(x))$ is continuous at $a$. $\blacksquare$
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License