In our last few lectures, we've focused on the limit of a function, which describes the behavior of a function near a point. Today, we ask a simpler question: what is happening at the point itself? When the limit of a function at a point and the function's actual value at that point coincide, we have a property called continuity. Intuitively, a continuous function is one whose graph has no breaks, jumps, or holes. You can draw it without lifting your pencil. ✏️
The intuitive idea of not lifting your pencil is a great start, but we need a precise, mathematical definition. This definition cleverly uses the concept of limits we've just mastered. A function $f$ is continuous at a number $a$ if it satisfies three specific conditions.
A function $f$ is continuous at a number $a$ if:
If any of these three conditions fail, the function is discontinuous at $a$.
Consider the piecewise function $f(x)$ defined below:
$$ f(x) = \begin{cases} \frac{x^2-4}{x+2} & \text{if } x < 1, x \neq -2 \\[1.5ex] 3 & \text{if } x = 1 \\[1.5ex] -x+3 & \text{if } 1 < x \le 4 \\[1.5ex] \frac{1}{x-4} & \text{if } x > 4 \end{cases} $$Let's find the numbers at which $f$ is discontinuous and explain why, using the three-part definition.
We can classify the different ways a function can be discontinuous. Understanding these types will deepen your grasp of limits and function behavior.
This occurs when the limit exists, but it is not equal to the function's value (either because $f(a)$ is defined as a different number, or is not defined at all). This creates a "hole" in the graph.
This typically occurs in piecewise functions where the graph "jumps" from one value to another. The left- and right-hand limits both exist, but they are not equal, causing the overall limit to fail.
This occurs where a function has a vertical asymptote. As $x$ approaches the point of discontinuity, the function's values grow without bound towards $\infty$ or $-\infty$.
A less common, but important, type of discontinuity occurs when a function oscillates so much near a point that it never approaches a single value. The classic example is $f(x) = \sin(1/x)$ at $x=0$.
We often want to talk about a function being continuous not just at a single point, but over its whole domain. Thankfully, most functions we are familiar with are continuous everywhere they are defined. This is a powerful result that follows from the Limit Laws.
The following types of functions are continuous at every number in their domains:
This means for these functions, evaluating a limit is as simple as evaluating the function. This is the Direct Substitution Property: if $f$ is a continuous function, then $\lim_{x \to a} f(x) = f(a)$ for any $a$ in the domain of $f$.
Find the limit: $\lim_{x \to 4} \frac{\sqrt{x} + 5}{x - 1}$.
The function is a quotient of functions that are continuous on their domains. We check if the point $x=4$ is in the domain. The numerator $\sqrt{x}+5$ is defined at $x=4$. The denominator $x-1$ is defined and is not zero at $x=4$. Since $x=4$ is in the domain, we can use direct substitution:
$$ \lim_{x \to 4} \frac{\sqrt{x} + 5}{x - 1} = \frac{\sqrt{4} + 5}{4 - 1} = \frac{2+5}{3} = \frac{7}{3} $$
Remember, the Direct Substitution shortcut only works because the value we are approaching, $a$, is in the domain of the function. If substituting $a$ results in division by zero or the square root of a negative number, you must use other techniques like factoring or conjugates, as we saw in the previous lecture.
If $f$ and $g$ are continuous at $a$ and $c$ is a constant, then the following functions are also continuous at $a$:
How can we discuss continuity at an endpoint like $x=3$ for a function like $f(x) = \sqrt{9-x^2}$, whose domain is $[-3, 3]$? For this, we use one-sided continuity.
A function $f$ is continuous from the right at $a$ if $\lim_{x \to a^+} f(x) = f(a)$.
A function $f$ is continuous from the left at $a$ if $\lim_{x \to a^-} f(x) = f(a)$.
Let's test $f(x) = \sqrt{9-x^2}$ at its endpoints:
$\lim_{x \to 3^-} \sqrt{9-x^2} = \sqrt{9-9} = 0$. And $f(3) = \sqrt{9-3^2} = 0$.
Since the limit from the left equals the function's value, $f$ is continuous from the left at $x=3$.$\lim_{x \to -3^+} \sqrt{9-x^2} = \sqrt{9-(-3)^2} = 0$. And $f(-3) = \sqrt{9-(-3)^2} = 0$.
Since the limit from the right equals the function's value, $f$ is continuous from the right at $x=-3$.This confirms that the function is continuous on the entire closed interval $[-3, 3]$.
If $g$ is continuous at $a$ and $f$ is continuous at $g(a)$, then the composite function $f \circ g$ given by $(f \circ g)(x) = f(g(x))$ is continuous at $a$.
This allows us to evaluate limits by "bringing the limit inside":
$$ \lim_{x \to a} f(g(x)) = f\left(\lim_{x \to a} g(x)\right) $$
On what interval is the function $h(x) = \frac{1}{\sqrt{x-2}}$ continuous?
This function is a composition of $f(x)=1/x$ and $g(x)=\sqrt{x-2}$. We must find the set of all $x$ that satisfy two conditions.
1. $x$ must be in the domain of the inner function, $g(x)$.
The function $g(x) = \sqrt{x-2}$ is defined when $x-2 \ge 0$, which means $x \ge 2$. The domain is $[2, \infty)$.
2. The output $g(x)$ must be in the domain of the outer function, $f(x)$.
The function $f(x)=1/x$ is defined for all numbers except $0$. So, we must find which value of $x$ causes $g(x)$ to equal 0.
$$ \sqrt{x-2} = 0 \implies x-2=0 \implies x=2 $$
This means we must exclude $x=2$.
Combining these two restrictions (x must be in $[2, \infty)$ AND $x \neq 2$), we find that the function $h(x)$ is continuous on the interval $(2, \infty)$.
Problem: On what intervals is the function $H(x) = \ln(\sin(x))$ continuous?
One of the most important theorems about continuous functions is the Intermediate Value Theorem. It has profound implications and practical uses, like proving that equations have solutions.
Suppose that $f$ is continuous on the closed interval $[a, b]$ and let $N$ be any number between $f(a)$ and $f(b)$, where $f(a) \neq f(b)$. Then there exists a number $c$ in the interval $(a, b)$ such that $f(c) = N$.
In simple terms, if a function is continuous from one point to another, it must take on every single value (or "height") between those two points. It can't skip any values.
Show that there is a root of the equation $4x^3 - 6x^2 + 3x - 2 = 0$ between 1 and 2.
Let $f(x) = 4x^3 - 6x^2 + 3x - 2$. Since $f$ is a polynomial, it is continuous everywhere. We evaluate the function at the endpoints of the interval $[1, 2]$:
$f(1) = 4 - 6 + 3 - 2 = -1$.
$f(2) = 32 - 24 + 6 - 2 = 12$.
We are looking for a root, which means we want a number $c$ such that $f(c) = 0$. Since $f(1) = -1$ (negative) and $f(2) = 12$ (positive), the value $N=0$ lies between $f(1)$ and $f(2)$. By the IVT, there must be at least one number $c$ in $(1, 2)$ such that $f(c)=0$.
Find the value of the constant $c$ that makes the function $f(x) = \begin{cases} cx^2 + 1 & \text{if } x \le 2 \\ 2x - c & \text{if } x > 2 \end{cases}$ continuous everywhere by following these steps:
On what interval is the function $h(x) = \frac{\ln(x)}{\sqrt{4-x^2}}$ continuous?
Use the continuity of trigonometric functions to evaluate the limit: $\lim_{\theta \to \pi/2} \sin(\tan(\cos(\theta)))$.
Continuity is a simple yet profound idea that connects the value of a function at a point with the function's behavior near that point. By formalizing the "no breaks or jumps" intuition with a limit-based definition, we can analyze functions with great precision. The Intermediate Value Theorem is our first major application of continuity, giving us a powerful tool to prove the existence of solutions. This concept of unbroken, predictable functions is the foundation upon which the theory of differentiation is built, which is exactly where we're headed next. 🚀