The Limit

This is often the most difficult definition for students to grasp, but it is also the one which is most useful for doing sophisticated mathematics. Here, I will immediately point out that the limit has what is often referred to as an overloaded definition. There are several related, but subtly different ideas all of which are captured in the notation:

$$ \lim_{x \rightarrow x_0} Y $$

Since we speak English, we are used to overloaded definitions. For instance, a plane is a geometric object, a carpenter’s tool, and a vehicle for flying from one destination to the next. We use context to disambiguate! If I say “I flew in a plane” you do not imagine that I climbed inside a carpenter’s tool and whizzed about in the air. We will focus first on the functional definition of the limit. We say the limit of $f(x)$ is L precisely if for every $\varepsilon > 0$ there exists a $\delta > 0$ such that whenever $|x – x_0| \lt\delta$ then $|f(x) – L| \lt\varepsilon$.

This definition has a noncommutative order to it. I specify a delta, you find an epsilon. In our first encounters with this definition in examples we will find that we tend to work backwards from an assumed result. This really only works well because these are pedagogical examples. Out in the wild world of real mathematics, you must hunt down epsilons based upon the deltas and they may not behave how you want them to.

Next, I’d like to note that L is a value in the range of the function. It is a y-value! So, let us consider an easy example: let $f(x) = x + 1$ and consider $\lim_{x\rightarrow2}f(x) = 3$. We’ll prove this from the definition of the limit.

Let $\delta = \varepsilon > 0$ then suppose

$|x – 2| \lt \delta \rightarrow$

$|x + 1 – 1 – 2| \lt \delta \rightarrow$

$|(x + 1) – 3| \lt \delta \rightarrow$

$|f(x) – 3| \lt \delta \rightarrow$

$|f(x) – 3| \lt \epsilon $

This does not seem to have bought us much. We can just substitute 2 into the above function and obtain the same result. Why bother? What does the limit give us? The answer is easy if we take a step back and the ramifications will be astounding in scope. The limit lets us consider values that our function approaches, but that may not be in the domain of the function! That still doesn’t seem like much, but it is the foundation of ALL OF CALCULUS. In fact, the limit is so powerful that usually if we are looking to extend ideas of calculus, the definition of the limit is the first thing we try to tinker with. You will see this in multivariate calculus and complex analysis.

Let’s consider the following example: $g(t) = \frac{t^2 -1}{t – 1}$ and we want to consider $\lim_{x \rightarrow 1}g(t)$. But wait? 1 is not even in the domain of $g(x)$!

Let $1 > \delta = \varepsilon > 0$ then suppose

$|t – 1| \lt \delta \rightarrow$

$|t + 1 – 1 – 1| \lt \delta \rightarrow$

$|(t + 1) – 2| \lt \delta \rightarrow$

$\left|(t+1)\cdot\frac{t-1}{t-1} – 2 \right| \lt \delta \rightarrow$

$\left|\frac{t^2 -1}{t – 1} – 2 \right| \lt \delta \rightarrow$

$|g(t) – 2| \lt \delta \rightarrow$

$|g(t) – 2| \lt \epsilon$

So we determined this limit is 2! This is not revolutionary news, but it’s compelling. Notice the additional assumption I had to tack on to this example to make it work. Here, I had to guaranty that delta was less than 1. Why? Where did I use that assumption?

Imagine a tiny, tiny ant climbing on our function. The ant can report to which she climbs at any moment, but cannot see ahead or behind herself. She can always start near our point of interest, but not at it. Moreover, she will fall through any hole or off any ledge, no matter how small. If she can crawl to our destination and report her height than the height she gets to is our limit.

True limits are two sided, her friend must be able to climb to the same height from the other side and they must be able to shake antennae. Again, they cannot shake antennae if there is a vertical gap, however small, because these ants are TINY! They can shake antennae if they get the vertical height, but there is exactly a missing point (known as a hole) in the graph.

In the diagram of $f(x)$ on the right, we see two ants approaching

$$\lim_{x\rightarrow-1}f(x)$$

from either side. The ants here, can get arbitrarily close and, as a result, can both come to the same height (roughly 3). Therefore,

$$\lim_{x\rightarrow-1}f(x) = 3$$

On, the other hand, what if I move the point of interest to -2, so that now, we are considering

$$\lim_{x\rightarrow-2}f(x)$$

Our black ant can scurry along $f(x)$ right up our point of interest and record her height as 0. But our orange ant has no place to start from! She must start to the left of the point, but here she cannot. Hence we say the limit from the right (indicated by the superscripted plus symbol) of $f(x)$ as x goes to negative two is equal to zero or

$$\lim_{x\rightarrow-2^+}f(x) = 0$$

but the limit from the left (indicated by the superscripted minus symbol) of $f(x) as x goes to negative two is undefined or

$$\lim_{x\rightarrow-2^-}f(x) = \varnothing$$

If the limit from the left does not agree with the limit from the right, then we say the limit does not exist. Hence

$$\lim_{x\rightarrow-2}f(x) = \varnothing$$

Consider this new function $g(x)$ in the third diagram on the right. Here, we see

$$\lim_{x\rightarrow0^+}g(x) = 1$$
$$\lim_{x\rightarrow0^-}g(x) = 0$$
$$\lim_{x\rightarrow0}g(x) = \varnothing$$

A final way of understanding limits is to use a tabular approach. It mirrors the geometric approach. This one has some serious drawbacks though in that it can lead you to false conclusion if you don’t have a clear understanding of the function you are dealing with. Let’s look at an example. Perhaps you have come upon the function $f(x) = \frac{sin(x)}{x}$ and you have little experience with trigonometric functions. You wish to know if the function has a limit at zero (who doesn’t?).