2.5: Continuity

Continuous Functions

In precalculus we learned that continuous functions are functions we can draw the graph of without lifting our pencil.

In calculus, use the previous sentence as intuition for the next definition:

A function $f$ is continuous at a number $x = a$ if the following conditions are satisfied:
  1. $f(a)$ is defined.
  2. $\displaystyle\lim_{x\rightarrow a}f(x)$ exists.
  3. $\displaystyle\lim_{x\rightarrow a}f(x) = f(a)$.

If any of these conditions fail, then the function is not continuous.

Note: The limit $\lim_{x\rightarrow a}$ doesn't see what happens at $x = a$.

Continuity is the reason why the limit does equal to the value at $a$.

Given this graph of a function $f$:

Where is the function not continuous and why?

In general:
  1. A function $f$ is said to be continuous on an interval if $f$ is continuous on every number in the interval.
  2. If $f$ is not continuous at $x = a$, it is said to be discontinuous at $x = a$.

To prove a function is continuous at $x = a$, you need to start with $\displaystyle \lim_{x\rightarrow a} f(x)$ and show that limit is equal to the evaluation $f(a)$.

Show $f(x) = x^3 + 3x^2$ is continuous at every real number.
Use the definition of continuity to find out where each of the following functions are discontinuous:
  1. $f(x) = \dfrac{x^2 - x - 2}{x-2}$
  2. $f(x) = \begin{cases}\dfrac{1}{x^2} & x \neq 0 \\ 1 & x = 0\end{cases}$
  3. $f(x) = \begin{cases}\dfrac{x^2-x-2}{x-2} & x \neq 2 \\ 1 & x = 2\end{cases}$
  4. $f(t) = \begin{cases} 0 & t < 0 \\ 1 & t \geq 0 \end{cases}$

Which Functions are Continuous?

Continuity allows us to find limits quickly. Let's see what functions are continuous.

A polynomial function of degree $n$ is a function of the form \[f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_2x^2 + a_1x + a_0\qquad a_n \neq 0\] where $a_n, a_{n-1},\dots, a_1, a_0 \in \mathbb{R}$.

A rational function is a function of the form \[r(x) = \dfrac{p(x)}{q(x)}\] where $p(x)$ and $q(x)$ are both polynomials.

If $f$ and $g$ are continuous at $a$ and $c$ is a constant, then the following functions are also continuous at $a$:
  1. $f + g$
  2. $f - g$
  3. $cf$
  4. $fg$
  5. $f/g$ if $g(a) \neq 0$
If you take facts 1 and 3, we can create polynomial functions. Including fact 5 gives us rational functions. So in general:
  1. A polynomial function $P(x)$ is continuous on any value of $x$.
  2. A rational function $R(x) = p(x)/q(x)$ is continuous on every value of $x$ where $q(x) \neq 0$.
This means if we take any limit (except for limit at infinity) for polynomial and rational functions, we can just plug in the value!
Find \[\lim_{x\rightarrow -2} \dfrac{x^3 + 2x^2 - 1}{5 - 3x}\]

Turns out most of the functions we know about are continuous.

The following functions are continuous at every number in their domains:
  1. polynomials
  2. root functions
  3. exponential functions
  4. rational functions
  5. trigonometric functions
  6. logarithmic functions
Find \[\lim_{x\rightarrow\pi}\dfrac{\sin x}{2 + \cos x}\]
Where is $\dfrac{\ln x + e^x}{x^2 - 1}$ continuous?
Find \[\lim_{x\rightarrow e} \dfrac{\ln x + e^x}{x^2 - 1}\]

How about compositions?

If $g$ is continuous at $a$ and $f$ is continuous at $g(a)$, then the composite function $f \circ g$ is continuous at $a$.
Where are the following functions continuous?
  1. $h(x) = \sin (x^2)$
  2. $F(x) = \dfrac{1}{\sqrt{x^2 + 7} - 4}$

Intermediate Value Theorem

You are driving to Lake Tahoe from San Luis Obispo.

Your position at time $t$ (in seconds) is determined by the function \[f(t) = 4t^2\] where $f(t)$ is in feet.

At time $t = 0$, you are at position $f(0) = 0$; you hasn't moved from your starting position.

At time $t = 10$, you are at position $f(10) = 400$. This means you are 400 feet away from your starting position after 10 seconds.

This function is continuous. Intuitively, this means your car cannot suddenly teleport from one location to another.

As such, if we were to ask "is there a time $t$ where $f(t) = 100$?", the answer must be yes. This is called the Intermediate Value Theorem.

Intermediate Value Theorem
Suppose $f$ is continuous on $[a, b]$ and let $N$ be any number between $f(a)$ and $f(b)$ where $f(a) \neq f(b)$. Then there exists $c$ in $(a, b)$ such that $f(c) = N$.
Existence of Zeros of a Continuous Function
If $f$ is a continuous function on a closed interval $[a,b]$ and $f(a)$ and $f(b)$ have opposite signs, then there exists $c \in (a, b)$ where $f(c) = 0$.
Let $f(x) = 4x^3 - 6x^2 + 3x - 2$.
  1. Show $f$ is continuous for all $x \in \mathbb{R}$.
  2. Show there exists at least one $c \in (1,2)$ where $f(c) = 0$.