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 $f$: 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 $a$. $\blacksquare$