Limits and Continuity

Continuity §

Conditions for Continuity §

A function is continuous on its domain if and only if the following conditions are true for all values in its domain. When testing the continuity of $f(x)$ at $x=a$, test the following conditions, in order:

1. (condition $i$) ${\lim }_{x\to a^{-}}f(x)={\lim }_{x\to a^{+}}f(x)$, therefore ${\lim }_{x\to a}f(x)$ exists
2. (condition $ii$) $f(a)$ exists
3. (condition $iii$) ${\lim }_{x\to a}f(x)=f(a)$

If all of these are true for $x=a$, then you can say that $f(x)$ is continuous at $x=a$.

If not all of these are true for $x=a$, $f(x)$ is not continuous at $x=a$. You’ll now have to determine the type of discontinuity that you’re dealing with.

Discontinuity Types §

Here’s a cheat sheet.

flowchart TD

A["lim(x → a) f(x) exists"]
A -->|True| B["f(a) exists"] -->|False| G
B -->|True| F["lim(x → a) f(x) = f(a)"] -->|False| G
G["Removable Discontinuity"]

A -->|False| E["Either side of the limit approaches +/- infinity or DNE"]
E -->|True| H["Infinite Discontinuity"]
E -->|False| I["Both sides of the limit exist but are not equal"]
I -->|True| J["Jump Discontinuity"]

The type of discontinuity can be pretty easily determined based on the first condition it does not meet. This is why it is important to evaluate all of the conditions in order, as opposed to jumping around.

Condition $i$ Discontinuities §

Recall condition $i$: $i$. ${\lim }_{x\to a^{-}}f(x)={\lim }_{x\to a^{+}}f(x)$, therefore ${\lim }_{x\to a}f(x)$ exists

There’s two usual suspects for why a limit is not continuous at $x=a$.

Suspect 1 §

${\lim }_{x\to a^{-}}f(x)\ne {\lim }_{x\to a^{+}}f(x)$,
BUT both ${\lim }_{x\to a^{-}}f(x)$ and ${\lim }_{x\to a^{+}}f(x)$ exist.

Take the following example, represented by the one-sided limits defined and graphed below.

$$\underset{x\to 2^{-}}{\lim }f(x)=3,\underset{x\to 2^{+}}{\lim }f(x)=0$$

Obviously, since both sides are not equal, the limit doesn’t exist.

However, neither of these one-sided limits approach an asymptote (aka, approach either infinity) like we saw in Suspect 1, you can see that the value visibly “jumps” when you approach from either side and try to continue.

This is an example of a Jump Discontinuity, where a finite jump occurs at the meeting point ($x=a$) of two parts of the graph.

Suspect 2 §

${\lim }_{x\to a^{-}}f(x)=\pm \infty$ or ${\lim }_{x\to a^{+}}f(x)=\pm \infty$
OR either ${\lim }_{x\to a^{-}}f(x)$ or ${\lim }_{x\to a^{+}}f(x)$ does not exist.

This last is pretty straightforward; basically, if condition $i$ fails and one of your one-sided limits has a big fat infinity, wow, very easy, very cool, you have an Infinite Discontinuity, where there’s an infinite gap between the meeting point of two parts of the graph; i.e., vertical asymptote.

Here’s a more proper example:

$$\underset{x\to 3^{-}}{\lim }f(x)=-\infty ,\underset{x\to 3^{+}}{\lim }f(x)=\infty$$

Since the gap between $-\infty$ and $\infty$ is, well, infinite, we’ve satisfied the condition for an infinite discontinuity at $x=3$. Don’t forget that the function is still not continuous at $x=3$ though, it just has an all-new, fancier name.

Condition $ii$ Discontinuities §

Onto condition $ii$:
$ii$. $f(a)$ exists

This one is… really straightforward. $f(a)$ is just the y-value that is located at $x=a$. Is there anything there? Congrats, you don’t even have to look at condition $ii$ discontinuities, you’re back in business and moving on to condition $iii$.

If it doesn’t exist, though, you’re moving into easy discontinuity land with the last one you have to know: a Removable Discontinuity, where $f(a)$ doesn’t exist. But this guy’s a little secretive, okay? There’s a bit more to know about him later, but we’ll get to that. First, here is your obligatory example:

$$\underset{x\to 2}{\lim }f(x)=4,f(x)=\frac{(x-2)(x+2)(x-1)}{x-2},f(2)=DNE$$

Where the red and blue lines meet at $f(2)$ aka $x=2$ aka “just pop the thing x is approaching in the equation point” is a hole. Let’s look at the equation a little more closely.

Because division exists, $(x-2)$ in the numerator can be divided out by $(x-2)$ in the denominator when solving for ${\lim }_{x\to 2}f(x)$, making it a valid limit approaching the location of (2,4), even if (2,4) doesn’t actually exist in the non-factored function. This is what makes this kind of discontinuity… well, removable! It is equal to $(x+2)(x-1)$ at all but one point that can be factored/removed out.

Condition $iii$ Discontinuities §

Okay, you know the drill by now. Here’s condition $iii$ if you are too lazy to scroll up: $iii$. ${\lim }_{x\to c}f(x)=f(c)$

Condition $iii$ is basically just a combination of the two steps above. Because we’ve gone in order, right… we know if we’re testing condition $iii$ that ${\lim }_{x\to a}f(x)$ exists and that $f(a)$ exists, so now we just have to make sure that they’re equal.

One of the most common places you’ll see this condition fail is in piecewise functions. Here’s a pretty simple example:

$$f(x)=\{\begin{array}{ll}{x}^2& \text{if }x\ne 1,\\ 2& \text{if }x=1\end{array}$$

Though ${\lim }_{x\to 1}f(x)$ exists, satisfying condition $i$, and $f(1)=2$ exists satisfying condition $ii$… since ${\lim }_{x\to 1}f(x)\ne 2$, we haven’t satisfied condition $iii$.

So now we’ve encountered our last discontinuity— wait. Wait. This guy’s a little familiar. It’s that shady Removable Discontinuity from earlier. How could that be?

Removable Discontinuities Logic §

Let’s consider the piecewise $f(x)$ above again, and bring back a concept from the condition $ii$ section.

Let’s define another function, $g(x)$, and check if it is continuous at $x=1$:

$$g(x)=\frac{x^2-1}{x-1}$$

First verify it through condition $i$.

$$\begin{array}{rlrl}\underset{x\to 1^{-}}{\lim }g(x)& =\frac{x^2-1}{x-1}& & \underset{x\to 1^{+}}{\lim }g(x)=\frac{x^2-1}{x-1}\\ \underset{x\to 1^{-}}{\lim }g(x)& =\frac{(x+1)(x-1)}{x-1}& & \underset{x\to 1^{+}}{\lim }g(x)=\frac{(x+1)(x-1)}{x-1}\\ \underset{x\to 1^{-}}{\lim }g(x)& =x+1& & \underset{x\to 1^{+}}{\lim }g(x)=x+1\\ \underset{x\to 1^{-}}{\lim }g(x)& =2& & \underset{x\to 1^{+}}{\lim }g(x)=2\end{array}$$

So, ${\lim }_{x\to 1^{-}}g(x)={\lim }_{x\to 1^{+}}g(x)$, which passes $i$. But $ii$ is where we find the issue, which you can probably tell by just looking at the equation…

$$g(1)=\frac{(1{)}^2-1}{(1)-1}$$

This, of course, yields a good old fashioned $\frac{0}{0}$, so $g(x)$ is discontinuous at $x=1$.

Why You Can't Divide By Zero

“You just can’t” is not an answer in Calculus, so let’s dive a little deeper and use an approach we’ve been picking at for a bit: one-sided limits. We will define $f(x)$ as follows:

$$f(x)=1/x$$

and to find the limit as $f(x)$ approaches $0$ from both sides, we will take a few numbers slightly smaller than $0$ (our ${\lim }_{x\to 0^{-}}$ side) and a few numbers slightly bigger than $0$ (our ${\lim }_{x\to 0^{+}}$ side), and put them in a table. This is probably one of the first methods you learnt to use to evaluate limits.

x	-0.1	-0.01	-0.001	-0.00001	0	0.00001	0.001	0.01
f(x)	-10	-100	-1000	-100000	?	100000	1000	100

So, from this table, we can evaluate both sides: ${\lim }_{x\to 0^{-}}f(x)=-\infty ,{\lim }_{x\to 0^{+}}f(x)=\infty$. This gives a limit that… does not exist, since both sides are not equal. Though there can be a point at a place where a limit does not exist, at $0$ we’re just between two asymptotes…

Because we got a limit that exists from condition $i$— which, I will stress again, you can only move onto condition $ii$ if that is the outcome of condition $i$, this is the most important thing in this entire continuity shebang— we know that it is possible to turn $g(x)$ into something that does give us a value that exists at $x=1$.

To make ${\lim }_{x\to 1}$ exist, we had to remove the hole at $(x-1)$ by cancelling out that value from the numerator and the denominator. This gave the equation $x+1$, with a domain— boo all you want, time for those algebra war flashbacks— of $(-\infty ,\infty )$, as opposed to the domain of our original, unmanipulated $g(x)$, which had a domain of $(-\infty ,1),(1,\infty )$.

But for every value that isn’t $1$ there’s a pretty obvious pattern that you can notice. $x+1$ is equal to $g(x)$ at every point, except for that troublemaker 1 (or $a$, for future reference vis-à-vis the notation we used to define our continuity conditions).

So, now that we’ve done all that work, what is a Removable Discontinuity? And why is $f(x)$— yes, from up there, all the way up there, I put it down here too so you don’t have to lift a finger— an example of one?

$$f(x)=\{\begin{array}{ll}{x}^2& \text{if }x\ne 1,\\ 2& \text{if }x=1\end{array}$$

We know that $i$ and $ii$ are satisfied, $iii$ is not because ${\lim }_{x\to 1}f(x)\ne 2$, we know that if we’ve meandered all the way to condition $iii$ we’ve stumbled upon this prima-donna discontinuity. We know that we can “fix” this discontinuity by factoring. How do you factor a piecewise though?

You don’t. A Removable Discontinuity can be “fixed” by defining a new function $g(x)$— a different $g(x)$ from what we know, same name different guy, Mike Tyson, baseballer versus Mike Tyson, Mike Tyson.

Is it really that easy? §

It is literally that easy. You “fix” a Removable Discontinuity by defining a new function, your $g(x)$, so that ${\lim }_{x\to a}g(x)=a$. You can’t just choose an arbitrary value to slide in here and forget about $f(x)$, though.

Let’s call whatever ${\lim }_{x\to a}f(x)$ does equal $L$. When we’re using $f(x)$ to create $g(x)$, we’re not modifying $f(x)$ itself— instead, we’re actually manipulating $L$.

$$g(x)=\{\begin{array}{ll}f(x)& \text{for }x\ne a,\\ L& \text{for }x=a,\end{array}$$

By going in and re-defining $f(x)$ as $g(x)$, you create a function that is equal to $f(x)$ at all but one point. Our shiny, all-new $g(x)$ is now continuous at $a$.

Sound familiar? You know what else sounds familiar? Mike Tyson. AKA, old $g(x)$. AKA, his name is now $h(x)$, because we need a bit of diversity in these un-ending function definitions.

$$h(x)=\frac{x^2-1}{x+1}$$

We’re still concerned with finding if this is continuous at $x=0$. We worked this out originally by simplifying our function in the algebraic way— factoring, which you’ve been doing since, what, the fourth grade? But let’s think outside of the box a little— think in terms of our last definition of $g(x)$ as it relates to ${\lim }_{x\to a}f(x)$.

Considering ${\lim }_{x\to 1}h(x)=2$, we’ll establish that $a$ = 1 and $L$ = $2$. Let’s create a new function, $h_2(x)$, that manipulates $L$ such that $h_2(x)$ is continuous at $x=1$.

$$h_2(x)=\{\begin{array}{ll}h(x)& \text{for }x\ne 2,\\ 2& \text{for }x=2,\end{array}$$

Boom. Continuous. And essentially the same as if we said $h_2(x)=x+1$! Because math.

So… removable discontinuities… §

Oh, yeah. I almost forgot.

Above all, a Removable Discontinuity is punishment for being annoying and not simplifying to lowest factors, or being a weirdo and actively trying to make everyone’s lives worse by creating arbitrary holes.

…but it’s also actually applicable and useful, I promise! Just not in a way that immediately comes to mind. All of our examples so far have dealt with polynomials, but at some point we’re going to need to take off the kiddie training wheels and dive into non-polynomial discontinuities. Scary, I know.