The first concept in calculus, and the one that underpins all the others, is the idea of a limit, so let's take at it. Suppose we have a function $f(x)$, and we would like to know what's happening at some $x$ value (call it $x=c$), but we can't actually evaluate $f(c)$ (it wouldn't really be an interesting question if we could).

For example, consider $f(x) = \dfrac{x^2-1}{x-1}$ with $c=1$. We can't actually evaluate $f(1)$, because that would cause us to divide by zero, breaking the cardinal rule of mathematics.

However, we also don't just throw our hands up and say that this function is a complete mystery at $x=1$; instead, we dig a bit deeper. You may be tempted, based on your algebra experience, to simplify the function by factoring and canceling an $x-1$ term: $\begin{aligned} f(x) &= \dfrac{x^2-1}{x-1}\ &= \dfrac{(x+1)(x-1)}{x-1}\ &= \dfrac{(x+1){\color{red} \cancel{(x-1)}}}{{\color{red} \cancel{x-1}}}\ &= x+1 \end{aligned}$

Aha! you say; this function is much simpler than it first appears: $f(x) = \dfrac{x^2-1}{x-1} = x+1$ and we can easily evaluate the simplified function at $x=1$, since we've cancelled the term that caused us to divide by zero earlier: $f(1) = 2.$

Is it quite that simple, though? We couldn't evaluate the function at $-1$, meaning that the domain didn't include $-1$, but after some algebra, we suddenly could. Did the domain change? What's going on here?

It doesn't help much to graph the two functions. If you use a calculator or computer graphing systems, you'll see the same picture whether you enter $\dfrac{x^2-1}{x-1} \textrm{ or } x+1.$

Here's what's happening: in the original function, there's a discontinuity at $x=1$; that doesn't belong to the domain. By canceling the $x-1$ term, we haven't affected the domain, but only the way that the function is written. What this means is that there's still a discontinuity there, but it's simply a hole in the graph at that one point; everywhere else, the graph looks exactly like $x+1$.

We can sum this up by saying that $\dfrac{x^2-1}{x-1} = x+1 \textrm{, as long as } x\ \cancel{=}\ 1.$

Taking the limit is like approaching this hole from either side, never quite reaching it, but going far enough to see a pattern. If an ant walked along the graph toward the place where $x=1$, the function values would get closer and closer to $2$, so we say that the limit of this function is $2$ as $x$ approaches $1$; we write this as $\lim_{x \to 1} \dfrac{x^2-1}{x-1} = 2$

Using a Table to Find a Limit

We've seen one way to find a limit, by using algebraic steps to simplify the function in question. Another way to answer the question is to think like the ant, crawling along the graph of the function. In other words, check the function value close to $x=1$ on either side and look for a trend. If we evaluate $f(x)$ at $0.9, 0.99, 0.999$, for instance, we should see the answers getting closer to $2$: $\begin{array}{c c} \begin{array}{c c} x & f(x)\ \hline 0.9 & 1.9\ 0.99 & 1.99\ 0.999 & 1.999 \end{array} & \begin{array}{c c} x & f(x)\ \hline 1.1 & 2.1\ 1.01 & 2.01\ 1.001 & 2.001 \end{array} \end{array}$

Notice that we're approaching $x=1$ from both sides, and in both cases, the function values get closer and closer to $2$ (never actually reaching $2$, but that's the idea of a limit).

Examples

Application: Rates of Change

Here's a preview of the next big topic, the one that will dominate most of this course: the derivative.

Where you see "rate of change," think of speed. For instance, if you were on a road trip and stopped at two rest stops $300$ miles apart, the first at 1:00 pm and the second at 6:00 pm, you'd know that your average speed between those two stops was $60$ mph, because you traveled $300$ miles in $5$ hours, and $\dfrac{300 \textrm{ miles}}{5 \textrm{ hours}} = 60\ \dfrac{\textrm{miles}}{\textrm{hour}}$

This is your average speed, but that doesn't mean you were always traveling at exactly $60$ mph on the trip; as you sped up and slowed down or stopped, it all evened out to an average of $60$.

Let's think about the average rate of change for a generic function. For example, consider $f(x) = x^2$, and find the average rate of change of this function from $x=1$ to $x=2$.

Think of the shift from $x=1$ to $x=2$ as the time elapsed, and the function as the position, to compare it to the traveling example.

At $x=1$, $f(x) = 1$, and at $x=2$, $f(x) = 4$.

${\color{red}\textrm{difference of } 1}\left{\begin{array}{c c c} x = 1 & : & f(x) = 1\ x = 2 & : & f(x) = 4 \end{array}\right} {\color{red}\textrm{difference of } 3}$

As $x$ (time) changed by $1$, $f(x)$ (position) changed by $3$, so the average rate of change is simply the difference in $f(x)$ divided by the difference in $x$: $\textrm{average rate of change } = \dfrac{3}{1} = 3$

Notice that the average rate of change is simply a new name for an old concept: we were computing the slope of the line that connects $(1,1)$ to $(2,4)$, the two points on the graph we were interested in; this line is called a secant line. ${\color{red}\textrm{average rate of change } = \dfrac{y_2 - y_1}{x_2 - x_1} = \textrm{ slope of secant line}}$

Average Velocity

If you throw a rock straight up from the ground at $96$ ft/s, the rock's height follows the path $f(t) = -16t^2 + 96t$

Here's a quick table with a few values; it'll be helpful to answer the question we'll encounter in a minute. $\begin{array}{c | c c c c c c c} t & 0 & 1 & 2 & 3 & 4 & 5 & 6\ \hline f(t) & 0 & 80 & 128 & 144 & 128 & 80 & 0 \end{array}$

Notice that it doesn't make any sense to go past $t=6$; the formula will happily give us a negative value, but this doesn't have any bearing on reality, where the rock will be lying inert on the ground.

Find the average velocity (rate of change of position) of the rock

1. from $t=1$ to $t=2$: $\dfrac{f(2) - f(1)}{2-1} = \dfrac{128-80}{2-1} = 48\ ft/s$
2. from $t=1$ to $t=3$: $\dfrac{f(3) - f(1)}{3-1} = \dfrac{144-80}{3-1} = 32\ ft/s$
3. from $t=3$ to $t=6$: $\dfrac{f(6) - f(3)}{6-3} = \dfrac{0-144}{6-3} = -48\ ft/s$

Notice that the average velocity is negative in the last case, because the rock is traveling downward.

Instantaneous Velocity

Okay, now how about the velocity of the rock at an instant in time? This is where we get to use what we learned earlier with limits.

Using the same example as before, we'd like to know the velocity of the rock at exactly $t=2$. How can we accomplish this? Using the process we just saw, we can find the average velocity between any two points in time, but can we use that to approach the problem of finding the instantaneous velocity at one moment in time?

It turns out that we can, by starting with the average velocity between $t=2$ and a moment soon after. For instance, the average velocity from $t=2$ to $t=2.1$ is likely to be at least close to the velocity at $t=2$. It's likely that the average from $2$ to $2.01$ will be an even better approximation, and from $2$ to $2.001$ better yet. Sound familiar? That's exactly what we were doing earlier with limits. If we could (theoretically) find the "average" velocity from $2$ to $2$, that would be the instantaneous velocity at $2$.

So if the average velocity from $2$ to another point $t$ nearby is $\dfrac{f(t)-f(2)}{t-2}$ the instantaneous velocity is $\lim_{t \to 2} \dfrac{f(t)-f(2)}{t-2}$

Let's try this out using a table like we did for the limit examples. Pick values for $t$ close to $2$ and evaluate this expression, looking for a trend.

$\begin{array}{c c} t & \dfrac{f(t)-f(2)}{t-2}\ & \ 2.1 & 30.4\ 2.01 & 31.84\ 2.001 & 31.984 \end{array}$

Since these values approach $32$, we conclude that at $t=2$ seconds, the instantaneous velocity of the rock is $32$ ft/s.

As it turns out, you've just done your first calculation of a derivative, the instantaneous rate of change. After we finish with limits, derivatives will occupy most of our time in this course, and we'll learn to do problems much more easily and quickly than we just did.

Technical Definition of a Limit

You can skip this section for now if you like; you'll see the precise mathematical definition of a limit, the way that professional mathematicians use it.

Let's start with our intuitive description of a limit, then move to an intermediate, slightly more precise description, before finally seeing the true definition.

Intuitive Description

We write $\lim_{x \to c} f(x) = L$ and say that the limit of $f(x)$ as $x$ approaches $c$ is $L$ if, as we trace along the graph of $f(x)$ by plugging in values close to $c$, the function values keep getting closer and closer to $L$.

Look back at the examples we did and see if you can connect the various pieces. For instance, in the first example, $f(x) = \dfrac{x^2-1}{x-1}$, $c=1$, and $L=2$.

Intermediate Description

We write $\lim_{x \to c} f(x) = L$ and say that the limit of $f(x)$ as $x$ approaches $c$ is $L$ if the values of $f(x)$ get arbitrarily close to $L$ if we make the values of $x$ close enough to $c$.

Think of it as a game: if you say that the answer is ($L$), I can challenge you to get within a distance of $0.1$ from $L$, or a distance of $0.0001$, or any distance I want (hence the arbitrary part), and you can do it, just by picking values of $x$ that are close enough to $c$.

Technical definition

To make the last leap, we'll introduce a couple of values, known as epsilon and delta (this is often called the epsilon-delta definition of a limit), and these will describe the process of getting arbitrarily close. Specifically, notice that, for instance, if we write $|x-c| < \delta$ we mean that the distance between $x$ and $c$ (the absolute value of their difference) is less than $\delta$, which is presumably some small number.

This looks complicated, but it's really saying the same thing as the intermediate description, except that it gives a very precise meaning to the concept of "arbitrarily close." If you like, you can study this to make sense of it, but it won't be crucial to our study of limits in this course.