7. Limits of Functions
≪ 6. Continuity | Table of Contents | 8. Derivatives and Local-to-Global Correspondences ≫Last week, we saw an introduction to continuous functions. Our heuristic was that a function \(f\) is continuous at a point \(x_0\) if \(x\approx x_0\) implies \(f(x) \approx f \left( x_0 \right)\). This is a statement about the “local” behaviour of \(f\): if you know what \(f\) looks like “near” \(x_0\), you can determine the value of \(f\left( x_0 \right)\).
Maybe another way to intuit this is, given a continuous function \(f\), someone annoying such as myself can erase a single point from the graph of \(f\). Using the property of continuity, however, one can easily “fill in” the hole that I created.
There is a way to make this idea rigorous, and that’s the notion of taking the limit of a function.
Definition 1.
Let \(f : D \to \mathbb{R}\) be a function defined on a subset \(D \subseteq \mathbb{R}\). Let \(x_0 \in \mathbb{R}\) be a point for which there exists at least one sequence \(\left( x_n \right) _{n\in \mathbb{N}} \subseteq D \setminus \left\lbrace x_0 \right\rbrace\) such that \(x_n \to x_0\).
The limit of \(f\) at \(x_0\), denoted \(\lim _{x\to x_0} f(x)\), is a real number \(L\in \mathbb{R}\) such that for all \(\epsilon > 0\), there exists a \(\delta > 0\) such that for all \(x\in D \setminus \left\lbrace x_0 \right\rbrace\) with \(\left\lvert x - x_0 \right\rvert < \delta \), \(\left\lvert f(x) - L \right\rvert < \epsilon \).
Equivalently, \(\lim _{x\to x_0} f(x) = L\) if for every sequence \(\left( x_n \right) _{n\in \mathbb{N}} \subseteq D \setminus \left\lbrace x_0 \right\rbrace\), \(\lim _{n\to\infty} f\left( x_n \right) = L\) (where this limit is that of a sequence).
Since there is some overlapping notation with limits of sequences, I’ll verbally distinguish “sequential limits” and “continuous limits”, the limits defined above. It should almost always be clear from context which is meant.
Immediately there is something clumsy and frankly unnatural about all the constraints on the domain \(D\) and the point \(x_0\) that we should try to elucidate: given \(f : D \to \mathbb{R}\), for which \(x_0\in \mathbb{R}\) can we even consider the quantity \(\lim _{x \to x_0} f(x)\)? Here I really want you to think about \(\lim _{x\to x_0} f(x)\) as asking the question, “If I erased the value of \(f\) at \(x_0\) (if it exists), are you able to fill it in?” To demonstrate what I mean, consider the domain \(D = \left\lbrace 0 \right\rbrace \cup (1, 2)\) and the function \[f(x) = \begin{cases} 0 & x = 0; \\ (x - 1) \sin \left( \frac{1}{(x-1)(x-2)} \right) & 1 < x < 2. \end{cases} \] You should observe the graph of this function on desmos.
With this function, consider the following five limits: \[\begin{align*} \lim _{x\to -1} f(x), && \lim _{x\to 0} f(x), && \lim _{x\to 1} f(x), && \lim _{x\to 1.5} f(x), && \textrm{and} && \lim _{x\to 2} f(x). \end{align*} \]
The limit \(\lim _{x\to -1} f(x)\) doesn’t make any sense: there is no information about \(f\) “near” the point \(x=-1\). This limit simply doesn’t exist.
For the next limit, it’s tempting to say that \(\lim _{x\to 0} f(x) = 0\). After all, \(f(0) = 0\), and that’s all we know about the function near the point \(0\). However, if I had erased the point \((0, 0)\) from the graph of \(f\), there’s no way to figure out what the value of \(f(0)\) “should” be! Thus, this limit doesn’t exist either for a reason similar to the previous example!
Now consider the limit \(\lim _{x\to 1} f(x)\). \(f\) is not defined at \(x = 1\), but visually it looks like its value should be \(0\) there. After all, the point \((1, 0)\) fits neatly onto the graph of \(f\).
The point here is that even though \(x=1\) is not in the domain of \(f\), it still makes sense to talk about the limit of \(f\) there!
Next, the limit \(\lim _{x\to 1.5} f(x)\) is simply \(f(1.5)\). One way to prove this is to use the fact that \(f\) is continuous at \(x=1.5\)… more on this later.
And finally, the limit \(\lim _{x\to 2} f(x)\). Similar to the limit as \(x\to 1\) and unlike the limits as \(x\to 0\) or \(x\to -1\), it makes sense to talk about this limit because the domain of \(f\) has lots of points that are “close to” \(x = 2\). But there’s visually no good way to fill in \(f(2)\) as a single point on the graph; it looks like there should be a whole line segment there. In this case the limit doesn’t exist.
Exercise 2.
Verify that \(\lim _{x\to 1}f(x) = 0\) and that \(\lim _{x\to 2} f(x)\) doesn’t exist by using the definition of a limit.
As with proving or disproving the continuity of a function, it’s generally easiest to prove a limit has a certain value by using \(\epsilon \)’s and \(\delta \)’s, and it’s generally easier to disprove taht a limit exists by using carefully constructed sequences.
Solution
To show that \(\lim _{x\to 1} f(x) = 0\), we let \(\epsilon > 0\). We wish to find some \(\delta > 0\) so that if \(x\in D\) and \(\left\lvert x - 1 \right\rvert < \delta \), \(\left\lvert f(x) - 0 \right\rvert < \epsilon \). As long as \(\delta < 1\), \(\left\lvert x - 1 \right\rvert < \delta \) implies \(x \neq 0\), so in particular \(f(x) = (x-1) \sin \left( \frac{1}{(x-1)(x-2)} \right)\). Therefore, \[\left\lvert f(x) - 0 \right\rvert = \left\lvert x-1 \right\rvert \cdot \left\lvert \sin \left( \frac{1}{(x-1)(x-2)} \right) \right\rvert \leq \left\lvert x - 1 \right\rvert < \delta .\] We conclude that if \(\delta = \min (\epsilon , 1)\), then for all \(x\in D\), \(\left\lvert x - 1 \right\rvert < \delta \) implies \(\left\lvert f(x) - 0 \right\rvert < \epsilon \). Since \(\epsilon \) was arbitrary, we conclude that \(\lim _{x\to 1} f(x) = 0\).
To show that \(\lim _{x\to 2} f(x)\) doesn’t exist, we first consider a sequence of points \(1 < x_n < 2\), \(x_n \to 2\) such that \[\frac{1}{\left( x_n - 1 \right) \left( x_n - 2 \right)} = 2 \pi n\] for all \(n\). Such a sequence morally speaking exists (you can solve for \(x_n\) explicitly by using the quadratic formula, if you really want to). In this case, \(f\left( x_n \right) = 0\) for all \(n\), so \(\lim _{n\to\infty} f\left( x_n \right) = 0\) (this limit is a sequential limit).
On the other hand, we can also find a sequence of points \(1 < x_n < 2\), \(x_n \to 2\) such that \[\frac{1}{\left( x_n - 1 \right) \left( x_n - 2 \right)} = 2 \pi n + \frac{\pi }{2}\] for all \(n\), in which case \(f\left( x_n \right) = x_n - 1\) for all \(n\). In particular, \(\lim _{n\to\infty} f\left( x_n \right) = 1\).
Thus there is no single value \(L\) that fits the sequential definition of the continuous limit, and we conclude that the desired limit doesn’t exist.
Continuing with the same function \(f\) defined on the same domain \(D\),
Exercise 3.
Is \(f\) continuous at \(0\)? Describe what needs to change about the following statement in order for it to be true:
A function \(f : D \to \mathbb{R}\) defined on a domain \(D \subseteq \mathbb{R}\) is continuous at \(x_0 \in D\) if and only if \(\lim _{x\to x_0} f(x) = f\left( x_0 \right)\).
The reason we emphasise these continuous limits so much and in particular need to think carefully about the domain of the function whose limit we’re taking is because of their applications to derivatives. Thinking ahead, we’ll be working with limits of the form \[\lim _{h\to 0} \frac{f(x+h) - f(x)}{h},\] and there’s no shot that the fraction in this limit is defined when \(h = 0\). Yet somehow we can make sense of this limit!
Question 4.
Going back to the definition of a continuous limit, what happens if we remove the restraints on \(x_0\)? What happens if we remove the restraint that \(x\in D \setminus \left\lbrace x_0 \right\rbrace\) and consider all \(x\in D\)? In particular, what changes about the above five limits?
Let’s quickly mention what happens when the symbol \(\infty\) appears. I’ll only mention the case of \(+\infty\), and you can hopefully imagine the case of \(-\infty\).
First, there’s the notion of a limit being infinite. In this case, it doesn’t make sense to write \(\left\lvert f(x) - \infty \right\rvert < \epsilon \), so we do something similar to sequences diverging to \(\infty\):
Definition 5.
Let \(f : D\to \mathbb{R}\) be a function with domain \(D \subseteq \mathbb{R}\). We say \(\lim _{x\to x_0} f(x) = \infty\) if for every \(M\in \mathbb{R}\), there exists \(\delta > 0\) such that for all \(x\in D\setminus \left\lbrace x_0 \right\rbrace\) such that \(\left\lvert x - x_0 \right\rvert < \delta \), we have \(f(x) \geq M\).
Equivalently, for every sequence \(\left( x_n \right) _{n\in \mathbb{N}} \subseteq D \setminus \left\lbrace x_0 \right\rbrace \) such that \(\lim _{n\to\infty} x_n = x\), one has \(\lim _{n\to\infty} f\left( x_n \right) = \infty\). (Both of these two limits are sequential.)
Question 6.
Write down a rigorous and precise definition of what it means for \(\lim _{x\to\infty} f(x) = L\), where \(L\in \mathbb{R}\cup \left\lbrace \infty \right\rbrace\).