# 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:

- 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:

- 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:

- 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:

- 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:

- 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:

- Therefore, $f \circ g = f(g(x))$ is continuous at $a$. $\blacksquare$