Defining a Limit Formally

After building the intuitive idea of what a limit is, you may be dissatisfied with the lack of rigor in how the concept is defined. Guessing the value of a limit like $\lim\limits_{x \to 2} \frac{(x +5 )(x - 2)}{(x - 2)} = 7$ is straightforward if you have an image of its graph, but how could you prove it? You could say “as $x$ approaches $2$, output values get closer to $7$”, but what does “closer” mean here? Do output values have to be strictly closing distance toward $7$? What if the output values trend towards $7$ but oscillate back and forth as $x$ approaches $2$?

Formalizing intuition

In order to prove that the limit equals $7$, we need to specify what a limit is. In the previous chapter, we looked at tables of $x$ and $y$ values, and if from both sides, as the $x$ value approached some given constant, the $y$ values appeared to converge somewhere, we concluded that the limit was at that point of convergence. The question that naturally appears is “How close to our posited point of convergence do we have to get in order to define the limit there?”

Let’s think back to our intuitive idea of the limit and see how we can apply it. Take $f(x)$ as the signum function, which returns an output of $-1$ for all negative inputs, $0$ for an input of $0$, and $1$ for all positive inputs. We want to test if $\lim_{x \to 0} f(x) = 0$ is true. As shown in Fig. 1, if you go any distance left or right of $x = 0$, you jump to $y = -1$ or $y = 1$. Even just by looking, this is uncontroversially a limit that does not exist, because the y-values are never “approaching” $y = 0$.

What would make this limit exist? Well, it might look like Fig. 2, where we smoothly curve the breakpoints into each other. By making it possible to go left/right of $(0, 0)$ by some infinitesimally small distance and not jump directly to $y = \pm 1$, the function becomes continuous. Therefore, in order to weed out these jumps, let’s create tiny distances from $y = 0$. If the limit exists, there should be a small window around the value our $x$ approaches where every single corresponding output is within that distance to $y = 0$. Otherwise, there would be a clear jump of output values immediately to the left or right of $x = 0$, just like with the signum function.

Since we’re still using the phrase “tiny distances”, let’s formalize that into a symmetric window around $y = 0$, as shown in Fig. 2 by the dashed lines. Call $\varepsilon$ (pronounced “epsilon”) the distance from the center to the top/bottom line, so in our case, we’d have a window created by the lines $y = 0 - \varepsilon$ and $y = 0 + \varepsilon$. Next, we need to make an appropriate window on the x-axis for any $\varepsilon$ value. Use the variable $\delta$ (pronounced “delta”) to make a window from $x = 0 - \delta$ to $x = 0 + \delta$. The criteria for a valid $\delta$ is that any $x$ you choose within that window will correspond to a $y$ within the window on the y-axis.

For an existent limit, this means that there’s a maximum $\delta$ value you could choose, as illustrated by the vertical lines in Fig. 2. If you try going higher than this value, one of the x-values near the edge of the window will correspond to a y-value outside our defined $\epsilon$ window. Anything lower will be valid, since it’d correspond to an output interval that’s a subset of our $\epsilon$ defined one. However, $\delta$ is never zero. Since the point of a limit is to see what the value approaches without actually taking into account the value itself, we exclude the x-value we’re approaching from the horizontal window test.

Let’s try to generalize this process for any function. We’ll use the generic form of $\lim_{x \to a} f(x) = L$. There’s $\varepsilon > 0$ for the vertical window, and we need a corresponding $\delta > 0$ such that if the distance of the $x$ value to $a$ is under $\delta$ but not zero (which can be notated as $0 < |x - a| < \delta$), then the distance from the output value to $L$ is less than $\varepsilon$ (notated as $|f(x) - L| < \varepsilon$). If we can show that a valid $\delta$ exists for any $\varepsilon > 0$, then $\lim_{x \to a} f(x) = L$ has been proven true.

Proving a limit statement

With our rigorous definition of a limit, let’s try proving limit statements. Remember that the goal is simply to prove that if $0 < |x - a| < \delta$, then $|f(x) - L| < \varepsilon$. If we can express $\delta$ in terms of $\varepsilon$, we can translate the right side of $|x - a| < \delta$ to something involving $\varepsilon$.

We’ll try to prove $\lim_{x \to 2} 2x = 4$. Although $2x$ is defined at $x = 2$, we’ll use this to get familiar with the structure of what an epsilon-delta proof looks like. Start by orienting ourselves with the goal: If $\varepsilon$ is a given positive number, find a $\delta$ such that if $0 < |x - 2| < \delta$, then $|2x - 4| < \varepsilon$. We factor $|2x - 4|$ to $2|x - 2|$. This changes our goal to proving $|x - 2| < \varepsilon / 2$.

Let’s set $\delta = \varepsilon / 2$ since our given was that $0 < |x - 2| < \delta$ and prove this works no matter what $\varepsilon$ was given. $|2x - 4| = 2|x - 2| < 2\delta$ just by doubling our given inequality. $2\delta = \varepsilon$ because of the $\delta$ we chose, so now we have $|2x - 4| < \varepsilon$. By taking our given conditions, we’ve proven the desired inequality, so $\lim_{x \to 2} 2x = 4$.

Our proof requires that we first express $\delta$ in terms of $\varepsilon$, because we are first given $\varepsilon$ and need to find a valid $\delta$ that works under those conditions. By finding a function with $\delta$ as the dependent variable, we can make it work for any $\varepsilon$ that exists. If you were wondering why we set $\delta = \varepsilon / 2$, it’s because we were reverse-engineering the desired result. We wanted $|2x - 4| < \varepsilon$, which algebraically can be manipulated to $|x - 2| < \varepsilon / 2$. Since we already had $|x - 2| < \delta$ from the original given, we wanted to make $\delta$ (a variable which we control) equal to $\varepsilon / 2$, since the proof becomes trivial after that step.

Since $\delta$ can be made smaller and still satisfy being in the $\varepsilon$ window (just a smaller subset of it), our proof still works if we tried setting $\delta = \varepsilon / 3$. Start with $|x - 2| < \delta$, then we have $|x - 2| < \varepsilon / 3$, so $2|x - 2| = |2x - 4|$ is less than $2\varepsilon / 3$, which still satisfies our original desired result of being less than $\varepsilon$.

Higher complexity

I designed the limit we just proved to be as straightforward as possible with delta-epsilon. We started by working backwards from the inequality we wanted to prove, then got the left side to be the same as the given inequality. After expressing $\delta$ in some form of $\varepsilon$, the proof could be finished simply by manipulating our given inequality into the desired inequality with clever substitution.

This is the extent of delta-epsilon you’d probably encounter in an introductory calculus course, but non-contrived limits are still provable. To finish off this chapter, let’s work through $\lim_{x \to 2} x^3 = 8$ and find a more general pattern. We want to prove for every positive $\varepsilon$, there is a $\delta$ such that if $0 < |x - 2| < \delta$, then $|x^3 - 8| < \varepsilon$.

Start by factoring $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$, which gets us $|x^3 - 8| = |x - 2| \cdot |x^2 + 2x + 4|$. This is where we branch off from the previous example, because $|x^2 + 2x + 4|$ has a variable inside it, so we can’t divide $\varepsilon$ by it. We’re going to try to constrain $\delta$ two different ways. We can set $\delta \leq 1$, then the quadratic reaches max at $x = 3$, giving us $|x^2 + 2x + 4| < 19$ for that defined $\delta$. Still with $\delta \leq 1$, $|x - 2| \cdot |x^2 + 2x + 4| < |x - 2| \cdot 19$, and our desired inequality has this less than $\varepsilon$, so we do $|x - 2| < \varepsilon / 19$.

In order for this to work, we needed both $\delta \leq 1$ for $|x^2 + 2x + 4| < 19$ and $\delta \leq \varepsilon / 19$ to get us to our desired inequality. Both will be satisfied if we just choose the smaller one, so use $\delta = \min(1, \varepsilon / 19)$. Now we’re at a familiar place, and the strategy you should remember for this general case is restricting $\delta$ one way and using its result to restrict it another way.

To finish off, we started with a given $\varepsilon > 0$, $0 < |x - 2| < \delta$, and we’ve defined $\delta = \min(1, \varepsilon / 19)$. We have $|x - 2| < 1$, and the $\delta \leq 1$ gives us $|x^2 + 2x + 4| < 19$. With our definition of $\delta$, we also get $|x - 2| < \varepsilon / 19$. Now, $|x^3 - 8|$ which becomes $|x - 2| \cdot |x^2 + 2x + 4|$ is less than $\varepsilon / 19 \cdot 19 = \varepsilon$, which proves the limit.

Now that we’ve formalized the limit, we can start becoming comfortable manipulating and composing limits, which is exactly what we’ll do in the next chapter. That is all for the epsilon-delta definition of limits, and until next time, I am out.