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Calculus Done Right
The mathematics of the continuous, through intuition not memorization.
In the last quiz, we looked at some examples of limits. Here is the general idea.
We say that the limit of the function \(f\) as \(x\) approaches \(a\) is the number \(L\) if, as \(x\) gets closer and closer to \(a,\) the function values \(f(x)\) get closer and closer to \(L.\) If there is no such number \(L,\) we say the limit does not exist.
When the limit exists, we use the notation \(\lim\limits_{x \to a} f(x) = L.\)
In this picture, for example, the limit of the function (in blue) as \(x\) approaches 2 (from either side) is 4.
As the input approaches 2, the output approaches 4.
In some ways this is a simple idea, but as we’ll see, there are plenty of subtleties involved!
Let’s start with a straightforward example. Here’s a graph of a function \(f.\) What is the limit of \(f(x)\) as \(x\) approaches 1? In other words, as the input gets closer and closer to 1, what value is the output getting closer to?
Now consider the function \(f\) given by the formula:
\[f(x) = \begin{cases}\begin{array}{rl} x^2, & x \neq 1 \\ 3, & x = 1 \\ \end{array}\end{cases}\]
In other words, \(f\) is the usual function \(y=x^2,\) except that we’ve set the value at \(x=1\) to be 3. What is \(\lim\limits_{x \to 1} f(x)\)?
In the previous example, the value of the function at 1 was 3. But the limit was still 1, because as the \(x\) values get closer and closer to 1, the function values get closer and closer to 1. This is an important point:
\(\lim\limits_{x \to a} f(x)\) has nothing to do with the value of \(f\) at \(a\) itself! It only says something about what happens as \(x\) gets close to \(a\).
Here’s another interesting example. Define
\[f(x) = \begin{cases}\begin{array}{rl} -1, & x \leq 0 \\ 1, & x > 0 \\ \end{array}\end{cases}\]
What is \(\lim\limits_{x \to 0} f(x)\)?
Note: For this to exist, the value of the function, \(f(x)\) must be getting closer to some number \(L\) as \(x\) gets closer to 0… no matter how close \(x\) gets to 0.
This is our first example in this quiz of a limit that doesn’t exist. It’s true that as \(x\) approaches 0 from the right, the function values approach 1. And as \(x\) approaches 0 from the left, the function values approach -1. But this means there’s no single \(L\) that the function approaches no matter how \(x\) gets close to 0. So the limit doesn’t exist.
This example, where the "right-hand" (as \(x\) approaches from the right) and "left-hand" (as \(x\) approaches from the left) limits exist but aren’t equal, is the simplest way a limit might not exist. But there are many other ways. For example, in the previous quiz we saw that \(\lim\limits_{x \to 0} \sin\left(\frac{1}{x}\right)\) does not exist, because as \(x\) gets small, \(\frac{1}{x}\) gets large, and so \(\sin\) just oscillates between -1 and 1, instead of approaching any particular \(L.\)
The function \(f(x),\) shown below, is defined on the interval \((0,10].\)
How many of the limits below exist?
\(\displaystyle \lim_{x \rightarrow 3} f(x)\)
\(\displaystyle \lim_{x \rightarrow 4} f(x)\)
\(\displaystyle \lim_{x \rightarrow 6} f(x)\)
\(\displaystyle \lim_{x \rightarrow 8} f(x)\)
Usually when we need to compute a limit in calculus, we won’t be presented with a graph, but with an algebraic expression. For example, let
\[f(x) = \frac{x^2+2x-8}{x-2}\]
What is \(\lim\limits_{x \to 3} f(x)?\)
The last example was easy, because everything was well-behaved at \(x=3.\) (In a later quiz, we’ll see this happens whenever the function is continuous.) Now consider:
\[\lim_{x \to 2} \frac{x^2+2x-8}{x-2}\]
The function is undefined at \(x=2,\) because of the denominator. We simply cannot evaluate \(f(2).\) But we can still investigate the limit as \(x\) approaches 2, because that only depends on what \(f\) is doing near 2, not at 2. In fact, notice that the numerator is also 0 when you plug in 2. This is another example of a \(\frac{0}{0}\) indeterminate form from the first chapter. When we encounter such a thing, the limit is not obvious. Often though, we can discover it by algebraic manipulation.
What is the limit? (Hint: factor the numerator.)
For the last three questions, we’re going to look at a strange and interesting example. The same basic idea behind limits hasn’t changed: \(\lim\limits_{x->a} f(x) = L\) means that as \(x\) approaches \(a,\) the values \(f(x)\) approach \(L.\)
Define:
\[f(x) = \begin{cases}\begin{array}{rl} x, & \text{if } \; x = \frac{1}{n} \text{ for some integer } n \\ 0, & \text{otherwise.} \\ \end{array}\end{cases}\]
For example, \(f(\frac{1}{2}) = \frac{1}{2},\) \(f(0) = 0,\) and \(f(\frac{2}{3}) = 0.\)
Try to get a feel for what this function looks like. There will be a picture on the next page, but see if you can work it out without looking.
The graph of \(f\) looks something like this:
The value of \(f\) is 0, except for at \(\frac{1}{n}\) points, which lie on the line \(y=x\).
Now, what is the limit of \(f(x)\) as \(x\) approaches \(\frac{3}{7}\)?
The value of \(f\) is 0, except for at \(\frac{1}{n}\) points, which lie on the line \(y=x\).
What about \(\displaystyle\lim_{x \to \frac{1}{3}} f(x)\ ?\)
The value of \(f\) is 0, except for at \(\frac{1}{n}\) points, which lie on the line \(y=x\).
Now for the most interesting question. What is \(\lim\limits_{x \to 0} f(x)\ ?\)
It’s hard to visualize exactly what’s happening near 0. But you know the rule: the function value is 0, except at points like \(\frac{1}{2},\) \(\frac{1}{3},\) \(\frac{1}{4},\) etc. So as \(x\) gets smaller and smaller, what happens to the \(f(x)\) values?