Limits: Exact Value or an Approximation?
Probably the majority of people will say that the limit is an exact construct and not an approximation but instead a unique real number. For the sake of simplicity we will keep it in the real number domain and use polynomials as examples. And indeed, this statement is correct. But books and intuitive explanations always talk about arbitrarily close etc. But arbitrarily close still is not exactly at the point right? It is just arbitrarily close ¯\_(ツ)_/¯
We will try to convince even the biggest skeptic (me) that the limit is indeed an exact number and not just an approximation.
For this, it might make sense to look at the derivative formula as it is intuitively relatively clear and makes up for great insights.
For all the mathematicians and rigorous folks, the following sections might sometimes not be completely accurate in favor of intuitive clarity. If you however spot a critical error, feel free to reach out!
Limit Notation
Stop, before we dive straight into derivatives we should make something clear. What the hell does \(\lim_{x \to a} f(x)\) mean? This expression reads as follows: “The limit f(x) as x approaches a”. To make it a bit less abstract, let’s say we have a function \(f(x)=x^2\) and we want to find its limit at \(x=3\). This leads to the following expression: \(\lim_{x \to 3} x^2\) which reads as “The limit of x^2 as x approaches 3”. The key here is that this expression does not simply set \(x=3\) and plug it into \(f(x)\). Instead, we approach \(f(x)\) from both sides which looks like the following table:
| Approach from the left | \(f(x)=x^2\) | Approach from the right | \(f(x)=x^2\) |
|---|---|---|---|
| 2.9 | 8.41 | 3.1 | 9.61 |
| 2.99 | 8.9401 | 3.01 | 9.0601 |
| 2.999 | 8.994001 | 3.001 | 9.006001 |
| 2.9999 | 8.99940001 | 3.0001 | 9.00060001 |
| 2.99999 | 8.9999400001 | 3.00001 | 9.0000600001 |
Table (1). Values approaching \(x=3\) from both sides for \(f(x)=x^2\)
As shown in Table (1), both sides approach 9 but they do not strictly equal 9 and on top of that we did not use \(x=3\) instead we used values slightly bigger or smaller than \(x\).
This also means that the limit and the actual function value at the point do not have to be the same thing. For example, we can define a function that behaves like \(f(x)=x\) everywhere around \(x=3\), but then manually set the value at \(x=3\) to 10:
$$ f(x)= \begin{cases} x, & x\neq 3 \\ 10, & x=3 \end{cases} $$
For now it suffices to understand, that the notation \(\lim_{x \to a} f(x)\), specifically the \(x\to a\) part, means that we approach something.
Derivative
Before we dive into the formalism of the derivative, let’s provide some intuition. Assuming we have well defined functions, the derivative provides the slope at any given point.
The idea behind the formula of the derivative is, that we take two points \(x_1\) and \(x_2\) on the x-axis and calculate the difference of their corresponding function values over the interval \([x_1,x_2]\).
We can do so by taking the difference of the values \(y_1=f(x_1)\) and \(y_2=f(x_2)\) and dividing them by the difference of the values \(x_1,x_2\). We can write this as follows:
$$ \begin{aligned} \text{(1)}\quad \frac{f(x_2)-f(x_1)}{x_2-x_1} = \frac{y_2-y_1}{x_2-x_1} \ = \frac{\Delta y}{\Delta x} \end{aligned} $$
Equation (1), provides the average slope over an interval \([x_1,x_2]\). In case you are unfamiliar or scared of \(\Delta x\) and \(\Delta y\), they are just a compact way, to write the differences between two points. The slope of equation (1) is also referred to as the slope of the secant line. This is simply the line we get, when drawing a line through \(y_1\) and \(y_2\).
While this sounds all super interesting and cool, we are still computing the slope between two different points, but the derivative provides the value at a single point. So how does this work?
In order to show this, let’s first dissect the formula for the derivative piece by piece. The definition of the derivative is the following:
$$ \text{(2)}\quad f’(x)=\lim_{h \to 0} \text{ }\frac{f(x+h)-f(x)}{h} $$
Now, let’s make this easier to digest and use \(f(x)=x^2\). When plugging in \(x^2\) we get the following:
$$ \begin{aligned} f’(x) &= \lim_{h \to 0} \text{ }\frac{(x+h)^2 - x^2}{h} \\ &= \lim_{h \to 0} \text{ }\frac{x^2 + 2xh + h^2 - x^2}{h} \\ &= \lim_{h \to 0} \text{ }\frac{2xh + h^2}{h} \\ &= \lim_{h \to 0} \text{ }2x + h \\ &= 2x \end{aligned} $$
After evaluating the expression we end up with \(f’(x)=2x\) which is also the derivative as we know. Previously, we were talking about secant lines and the derivative now provides us with the slope of the tangent line at one point. The difference between a secant and a tangent line is that a secant line is determined by two distinct points on the curve, while the tangent line captures the limiting slope at a single point. Thus, the derivative gives us the exact instantaneous slope at that point instead of an average slope over a small interval \([x_1,x_2]\).
But stop. The derivative formula above somehow bites the intuitive explanation that we approach something but do not set the \(x\) to the value we approach and thus in this case \(h\) never truly equals \(0\) but instead we approach \(0\). So where the hell did the \(h\) go? Why did it simply disappear if we said it never equals \(0\)? Well, before we can truly answer this question we must stay strong and dive just a tiny bit deeper.
The big ol’ \(\epsilon\)-\(\delta\)
The \(\epsilon\)-\(\delta\)-definition looks like this:
$$ \text{(3)}\quad \lim_{x \to a} f(x) = L \quad \Longleftrightarrow \quad \forall \epsilon > 0\ \exists \delta > 0 \forall x: 0 < |x-a| < \delta \Rightarrow |f(x)-L| < \epsilon $$
Yikes, that’s heavy. Let’s walk through it step by step. First of all we have an equivalence which means that from the right side the left side follows and vice versa. Let’s focus on the right part:
$$ \forall \epsilon > 0\ \exists \delta > 0 \forall x: 0 < |x-a| < \delta \Rightarrow |f(x)-L| < \epsilon $$
This is a conditional statement and reads as follows:
“For all epsilon bigger than 0, there exists a delta greater than 0, such that IF the distance between \(x\) and \(a\) is strictly greater than 0 and less than delta THEN the difference between \(f(x)\) and \(L\) is less than epsilon.”
Even in its written form it still sounds a bit heavy so let’s make use of a picture. We all know the saying “A picture tells a thousand words” ✨
Figure (1). \(\epsilon\)-\(\delta\)-Definition visualized. Source (modified): Math StackExchange
In Figure (1) there are two important parts. First the horizontal lines around the limit \(L\), spanning the red tube, which are denoted with \(L+\epsilon\) and \(L-\epsilon\). They set the error or maximum accepted deviation distance around the limit \(L\), that we accept for our function values \(f(x)\). In short, all values \(f(x)\) must fall within this range.
The second part shows a blue tube around \(a\) which is bounded by \(a+\delta\) and \(a-\delta\). This interval sets the boundary around the point \(a\) in which all points \(a_i\in [a-\delta;a+\delta]\) must satisfy \(|f(a_i)-L| < \epsilon\).
We can see that if we tighten the accepted error (red tube) we also have to tighten the blue tube because otherwise there may be inputs \(a_i\in [a-\delta;a+\delta]\) where \(|f(a_i)-L| \geq \epsilon\) and thus the condition is broken.
Putting it all together
Okay, we have now talked about derivatives and the \(\epsilon\)-\(\delta\)-definition. How should this convince even the biggest skeptic that limits are indeed exact? The ✨magic✨ is buried inside the \(\epsilon\)-\(\delta\)-definition and can be greatly understood with derivatives.
When going back to derivatives, we learned that the derivative \(f’(x)\) is the slope of the tangent at point \(x\). Furthermore, it is the value that all secants converge to, when the distance of the two points of the secants approach \(0.\) Looking at Figure (2), every secant slope is still an approximation because \(h\neq0\), but the value of the slope of those secants are forced toward the exact tangent slope which in turn is the limit.
And here comes the important part: In formula (2) we asked why \(h\) suddenly disappeared. The important point is that \(\lim_{x\to a}\) is an infinite process in which we get closer and closer but we never set \(h=0\) into the original fraction.
$$ \begin{aligned} &= \lim_{h \to 0} \text{ }\frac{2xh + h^2}{h} \\ &= \lim_{h \to 0} \text{ }2x + h \\ \end{aligned} $$
The original expression is even undefined at \(h=0\). We were only able to make this cancellation because we approach \(0\) but do not set \(h=0\). Instead, we simplified the expression for all \(h\neq0\).
This infinite process is stated in the \(\epsilon\)-\(\delta\)-definition in (3) which states that the right side of the equivalence must hold \(\forall\epsilon\) . This is also what we mean with infinite process, since there are infinitely many (positive) numbers and we can make the red tube in Figure (1) as small as we want. The condition \(0 < |x-a| < \delta\) describes exactly this idea, where we look for inputs arbitrarily close to \(a\) but not equal to \(a\) itself.
We can illustrate this behavior in Figure (2) where we can make the distance \(h\) smaller and smaller but for non-zero \(h\), we would get a secant and not a tangent. However, we are interested in the slope at the exact point and that is given by the tangent at that point. That requires us to take the limit, as we cannot just put \(h=0\) and magically let \(h\) disappear.
Another way to intuitively think about it is that the secant slopes approach one exact value, and because of that, the value at the exact point should behave like the slope of the tangent line at that point.
This further illustrates that the limit is a separate object from the value at \(h=0\). This is further illustrated in Figure (3) which shows a piecewise function where \(f(x)=x\) for all values except for \(x=3\). Now we see that \(\lim_{x\to 3} f(x)\) is equal to \(3\) but \(f(3)=10\). Therefore the limit when approaching \(a\) can be different from \(f(a)\) and hence the limit and the function value are two separate objects.
Since \(\lim_{x\to 3} f(x)=3\), the nearby behavior at \(x=3\) suggests that \(f(3)\) would be \(3\). However, we see that this is not the case and \(f(3)=10\). And since \(\lim_{x\to 3} f(x)=3\) (our expected behavior at \(x=3\)) was different from the actual function value \(f(3)\) we can go even further and deduce that the function is not continuous 💡.
But what does continuity mean? That’s another story :)
The End
Now you might think, what a wall of text, for something that is accepted as a fact. That is correct, but sometimes especially when working with infinities, it feels correct and wrong at the same time. It took me a while to intuitively accept the fact that limits are really exact and not just a super close approximation. And even when it made intuitive sense after working with limits a while, as soon as I stopped working with them and came back after a while, there was this weird feeling again. The explanation above however, satisfied my intuition. I think sometimes there are topics, where you need a certain amount of exposure in order to get this satisfied feeling. Another contributing factor is to really understand and dissect definitions. For a while I somewhat glossed over the \(\forall\epsilon\) requirement in the \(\epsilon\)-\(\delta\)-definition but this is the most crucial part, that also makes everything less magical. If this still does not convince you directly, that’s okay. Read the article again, take your time and do not hesitate to incorporate other resources.
With this post, I tried to provide different perspectives of the exactness of limits to satisfy other peers that might have the same weird feeling I once had.