What Is Continuity?

Take a look at the following two functions:

We say that a function $f(x)$ is continuous at a point $x = a$ if $f(x)$ is close to $f(a)$ for all values of $x$ close to $a$. For example, for the functions above:

$f(x)$ is continuous at $x = 2$, since for all values of $x$ close to 2, $f(x)$ is close to -1. In particular, $f(2.1) = -0.91$, and $f(1.99) = -1.01$, which are close to -1.

$g(x)$ is not continuous at $x = 2$, since for some values of $x$ close to 2, $g(x)$ is not close to 2. In particular, $g(1.9) = 0.52$, which is not close to 2.

Now we have to ask, what do we mean when we say that $f(x)$ is close to $f(a)$, or when we say that $x$ is close to $a$? Our goal now is to make our definition clear, and we'll go through a few attempts that don't work before we find one that does.

Attempt #1: $f(x)$ is continuous at a point $x = a$ if $f(x)$ is within 0.1 of $f(a)$ for all values of $x$ within 0.1 of $a$.

Here's an explanation:

Since $f(x) = x$, this definition requires $x$ to be within 0.1 of 0 for all values of $x$ within 0.1 of 0, which is clearly true. Therefore, under this definition, $f(x)$ is continuous at $x = 0$.

Since $g(x) = 10x$, this definition requires $10x$ to be within 0.1 of 0 for all values of $x$ within 0.1 of 0, which is not true; for example, if $x = 0.1$, then $10x = 1$, which is not within $0.1$ of $0$. Therefore, under this definition, $g(x)$ is not continuous at $x = 0$.

But $g(x) = 10x$ looks continuous at $x = 0$:

So this definition doesn't work. We might try to fix it by adjusting one of the 0.1's, like this:

Attempt #2: $f(x)$ is continuous at a point $x = a$ if $f(x)$ is within 0.1 of $f(a)$ for all values of $x$ within 0.01 of $a$.

So this definition doesn't work either. We'll have to do something a little more clever; here's our next attempt:

Attempt #3: $f(x)$ is continuous at a point $x = a$ if we can find a number $\delta$ such that $f(x)$ is within 0.1 of $f(a)$ for all values of $x$ within $\delta$ of $a$.

($\delta$ is a variable; it's the lowercase Greek letter delta.)

It may look like we've fixed our problem, but there's still an issue with our definition. Consider the following function:

$f(x) = \left\{ \begin{array}{lr} x/2 + 3 &\text{, }x < -1\\ x/2 + 3.06 &\text{, }x \ge -1 \end{array} \right.$

This function has a tiny jump at $x = -1$:

Our definition says this function is continuous at $x = -1$, since whenever $x$ is close enough to $-1$, $y$ is within 0.1 of 2.56, even if $x$ is to the left of the jump. But, this function just doesn't look like it should be continuous! So, how should we change our definition?

Here's the standard definition; see how yours compares:

Definition: $f(x)$ is continuous at a point $x = a$ if, for all $\varepsilon > 0$, we can find a number $\delta$ such that $f(x)$ is within $\varepsilon$ of $f(a)$ for all values of $x$ within $\delta$ of $a$.

($\varepsilon$ is also a variable; it's the lowercase Greek letter epsilon)

It may be possible to test if a graph is continuous using limits, specifically two-sided limits.

Here is a conjecture:

Suppose we have a graph $f(x)$ in which we want to test its continuity. If $$\lim_{x\to n^{\textbf{+}}}f(x)=\lim_{x\to n^{\textbf{-}}}f(x)$$ for all $n\in [a,b]$, then $f(x)$ is continuous in the range $[a,b]$.

This seems to be reasonable. For example, inspect the graph $f(x)=\dfrac{x^2}{x^2-9}$ below:

We see that if we take the limit from both sides anywhere except $x=\pm 3$, then it turns out that $\displaystyle\lim_{x\to n^+}f(x)=\displaystyle\lim_{x\to n^-}f(x)$. When $x=-3$, this function is not continuous. And just as we thought, $\displaystyle\lim \limits_{x \to n^+}f(x)=-\infty \text{ while } \displaystyle\lim_{x\to n^-}f(x)=\infty$, which are not equal. Ditto for $x=3$.

Now let's look at the graph $f(x)=\dfrac{1}{x^2}$:

We can see that the graph has $\displaystyle\lim_{x\to n^+}f(x)=\displaystyle\lim_{x\to n^-}f(x)$ for all $n$; however, there is a discontinuity at $n=0$, yet $\displaystyle\lim_{x\to 0^+}f(x)=\displaystyle\lim_{x\to 0^-}f(x)=\infty$.

Thus, our method only works for certain essential discontinuities.

However, this is just essential discontinuities. There is also another type of discontinuity we have not been over: removable discontinuities.

Let's see what happens when we try to use our conjecture on removable discontinuities:

Consider the graph $f(x)=\dfrac{x}{x^2-4x}$ show below.

We see that there is a removable discontinuity at $x=0$. However, note that $$\displaystyle\lim_{x\to 0^+}f(x)=\displaystyle\lim_{x\to 0^-}f(x)=-\dfrac{1}{4}$$

Thus, our method doesn't work for removable discontinuities.

(Videos) Continuity and Limits Made Easy

by PatrickJMT

Here's a video by patrickJMT explaining what continuity is and the different types.

A function is continuous if you can draw it without lifting your pen or pencil. If there's a hole or a break in the graph (like in the picture below) you have to hop over it to go to the next part.

Suppose we have a function $f(x)$ which is graphed below. Beneath the circle the $x$-coordinate is $a$ and the $y$-coordinate is $L$. Also suppose we have a dot shaded in above and its $y$-coordinate is $M$.

If we take the limit of our function $f(x)$ as $x$ approaches $a$, we see that the points get closer and closer to our whole which is $L$. So, $$\lim_{x\rightarrow a}f(x)=L$$ If we were to plug $a$ into our function though, it would take us to the dot above at $M$. So, $$f(a)=M$$ Clearly, $$\lim_{x\rightarrow a}f(x)\ne f(a)$$

If we wanted to make the function continuous all you have to do is move the point into the circle to have it filled in.

Our function is now continuous because we can trace its path without having to stop. Now, the function at the point $a$ is equal to $L$: $$f(a)=L$$ And this is now equal to the limit at that point, $$\lim_{x\rightarrow a}f(x)=f(a)$$ This becomes our definition.

A function $f(x)$ is continuous at $x=a$ if $\lim_{x\rightarrow a}f(x)=f(a)$.

Types of Discontinuity

There are three different types of discontinuity.

1. Removable Discontinuity
2. Infinite Discontinuity
3. Jump Discontinuities.

A removable discontinuity is the kind we saw above. There's a single point removed leaving a hole.

Algebraically we can tell this because the limit exists but the value at that point is different. $$\lim_{x\rightarrow a}f(x)=L;\ \ f(a)=M$$

An infinite discontinuity is when the function spikes up to infinity at a certain point from both sides.

Algebraically we can tell this because the limit equals either positive infinity or negative infinity. $$\lim_{x\rightarrow a}f(x)=\pm\infty$$

A jump discontinuity is when the function jumps from one location to another.

Algebraically we can tell this because the limit at $a$ does not exist. The limit from either side of $a$ goes to two different points. $$\lim_{x\rightarrow a^-}f(x)=M;\ \ \lim_{x\rightarrow a^+}f(x)=N$$

Here's another video on the topic of continuity with some examples.

The following functions are continuous on their domains:

1. Polynomials
2. Rational Functions
3. Trig Functions
4. Root Functions

The example solves is finding any discontinuities for the function $$f(x)=\begin{cases}x^2+3x&x<0\\\sqrt{x}+1&x\ge0\end{cases}$$ The first function is a polynomial so we know that it's continuous on its interval. The second function is a root function so we know it too is continuous on its interval. The only place where there might be a discontinuity is at $x=0$.

To figure this out we need to check to see if the limit at zero equals the value at zero. $$\lim_{x\rightarrow0}f(x)≟f(0)$$ To find the value at $0$ we use the second function because that includes zero. $$\sqrt{x}+1=\sqrt0+1=1$$ So now we have $$\lim_{x\rightarrow0}f(x)≟1$$ Next, we need to figure out what the limit is equal to. To determine this we check the limit at each side of zero. $$\lim_{x\rightarrow0^-}\text{ and }\lim_{x\rightarrow0^+}$$ Since the first function includes values to the less than zero we use it to find the left of zero limit and since the second function includes values greater than zero we use it to find the right of zero limit. $$\lim_{x\rightarrow0^-}x^2+3x\ \text{ and }\lim_{x\rightarrow0^+}\sqrt x+1$$ To find these limits we just plug zero into the respective function. $$\lim_{x\rightarrow0^-}x^2+3x=0\\ \lim_{x\rightarrow0^+}\sqrt x+1=1$$ As you can see, these two values are not equal so therefore the limit doesn't exist. This tells us that we have a jump discontinuity at $x=0$. If we graph this we can see it easily.

(Video) Continuity on Piecewise Functions

by PatrickJMT

Here's a video by patrickJMT showing you how to figure out where a piecewise function is not continuous.

While you can always graph a function to see if there's any discontinuities, it can take awhile to do so. In this video Patrick shows you how you can find discontinuities without graphing.

He makes use of the definition of continuity and a theorem. First, we say $f(x)$ is continuous at $x=a$ if $\lim\limits_{x\rightarrow a}f(x)=f(a)$. The theorem he uses states:

Trigonometric, rational root, and polynomial functions are continuous on their domains.

The question he looks at is: At what points, if any, is the following function NOT continuous? $$g(x)=\begin{cases}x-3;&x\le-1\\x^2+1;&-1<x\le2\\x^3+4;&x>2 \end{cases}$$

First look at the individual functions in the piecewise. Notice that they're all just polynomials. According to our theorem, these are all continuous on their domains. Since these are continuous everywhere the only potential places for discontinuity is where they switch from one function to another; at the breaks.

If they switch at the same point, then the piecewise function is continuous at that point. If they have two different values then this means there's a jump discontinuity happening at that point.

So let's see if there's any discontinuities in this piecewise function. The first break we have is at $x=-1$. We can use the limit definition to see if the limit is equal from both sides. $$\lim_{x\rightarrow-1^-}g(x)≟\lim_{x\rightarrow-1^+}g(x)$$ If the limits are equal we can then figure out if there's continuity at the point using the definition of continuity. $$\lim_{x\rightarrow-1}g(x)≟g(-1)$$

So let's find the limit from each side. $$\lim_{x\rightarrow-1^-}g(x)=\lim_{x\rightarrow-1^-}{x-3}\\ =\color{red}{(-1)}-3=-4$$

So the limits from each side are not equal. Therefore, the limit at $-1$ does not exist. So the function is not continuous at $x=-1$. $$\lim_{x\rightarrow-1^-}g(x)\ne\lim_{x\rightarrow-1^+}g(x)\\ \Rightarrow\lim_{x\rightarrow-1}g(x)\text{ DNE}\\ \Rightarrow\lim_{x\rightarrow-1}g(x)\ne g(-1)$$

The second break is at $x=2$. We do the same process to figure out if it's continuous there. $$\lim_{x\rightarrow2^-}g(x)=\lim_{x\rightarrow2^-}{x^2+1}\\ =\color{red}{(2)}^2+1=5$$

So the limits from each side are not equal. Therefore, the limit at $2$ does not exist. So the function is not continuous at $x=2$. $$\lim_{x\rightarrow2^-}g(x)\ne\lim_{x\rightarrow2^+}g(x)\\ \Rightarrow\lim_{x\rightarrow2}g(x)\text{ DNE}\\ \Rightarrow\lim_{x\rightarrow2}g(x)\ne g(2)$$