# Introduction to Limits and Derivatives

Overview of the derivative

In the previous lesson, we discuss how when $$f'(x)$$ is a constant, we can go from $$f(x)$$ to $$f'(x)$$ and from $$f'(x)$$ to $$f(x)$$ by finding the slope and area, respectively. The "calculation procedures" we used to find the slope and area were pretty simple. When $$f'(x)$$ is changing (that is, when $$f'(x)$$ is not a constant), we still have to find the slope and area to go back and forth between between $$f(x)$$ and $$f'(x)$$; but we'll see that we'll have to use a fairly sophisticated calculation procedure to get the slope and the area. For about the first half of our studies of calculus, we're going to be studying differential calculus: how to go from $$f(x)$$ to $$f'(x)$$. In this lesson we're going to discuss something called a derivative. The derivative of $$f(x)$$ is written as either $$f'(x)$$ or $$df/dx$$. To go from $$f'(x)$$ (or $$df/dx$$) to $$f(x)$$, we must take the integral of $$f'(x)$$ to get the area underneath $$f'(x)$$. This integral is written as

$$\int{f'(x)}dx=f(x).$$

We're going to learn later that both of these notations involving the '$$d$$' letters and the elongated s-shape are very appropriate. You can already see, albeit on a superficial level, that $$df/dx$$ looks kind of similar to $$Δf/Δx$$ and $$\sum{f'(x)Δx}$$ looks similar to $$\int{f'(x)}dx$$—but we'll get in all of that later.

In Figure 1, I have drawn a graph of some arbitrary function $$g(x)$$ which could be, pretty much, anything.$$^1$$ Before discussing how to calculate $$g'(x)$$ given $$g(x)$$, let's start out by discussing what a derivative is. In a nutshell: the derivative $$g'(x)$$ is just the slope of $$g(x)$$ at each point. We can see, qualitatively, that the steepness of $$g(x)$$ within the whereabouts of the point $$(x_1,g(x_1))$$ looks pretty flat whereas the steepness around the point $$(x_2,g(x_2))$$ is quite steep.

In order to calculate $$g'(x)$$ from $$g(x)$$ at each point along the curve, we'll have to introduce a new concept known as a limit which is perhaps the most important idea in calculus since it makes taking derivatives and integrals possible.

Figure 1: Above, I have graphed some arbitrary function $$g(x)$$. At the point $$g(x_0)$$, the slope or steepness of the curve is very small (indeed, it is zero) whereas at the point $$g(x_1)$$ the slope of the curve is comparitevely large.

Basic example of a limit

"Introduction to the intuition behind limits."$$^{}$$

Below, I have drawn a graph of the function $$f(x)=x^2$$ (see Figure 2). As the value of $$x$$ gets closer and closer to $$2$$ (which we write as $$x→2$$) and as $$Δx$$ gets closer and closer to zero (or, a shorter way of saying that: as $$Δx→0$$), what does the value of $$f(x)=x^2$$ get closer and closer to equaling? That entire sentence can be rewritten as

$$\lim_{Δx\to0}x^2=?\tag{1}$$

Equation (1) is read as follows: "the limit of $$x^2$$ as $$Δx$$ approaches $$0$$ is equal to $$?$$" Equation (1) should be interpreted like this: as we keep picking values of $$Δx$$ which get closer and closer to equaling zero (which, to restate in another way to make sure that we're on the same page; as we keep picking values of $$x$$ that get closer and closer to equaling $$2$$), what are the values of $$x^2$$ getting closer and closer to equaling? What goes on the right-hand side of Equation (1) is the thing that $$x^2$$ is getting closer and closer to equaling (I just put a $$?$$ mark on the right-hand side since we're not sure yet what that thing is).

Table 1: Values of $$f(x)$$ as $$x→2$$ and as $$Δx→0$$.

This is fairly straightforward and can be solved through repetition by calculating Equation (1) for several different values of $$x$$ which get successively closer and closer to equaling $$2$$. Since we want to choose values of $$Δx$$ that are close to zero, let's choose $$Δx=0.1$$. This means that $$x=2.1$$ and $$x^2=4.41$$. To find the answer to Equation (1), we want to keep choosing values of $$Δx$$ that keep getting closer to zero. Let's pick $$Δx=0.01$$. In this case, we have $$x=2.01$$ and $$x^2=4.0401$$. Ok, let's now pick a value of $$Δx$$ that's even closer to being zero; let's say $$Δx=0.001$$. Then, $$x=2.001$$ and $$x^2=4.004001$$. For convenience, I have listed all of the calculations we just did in Table 1 on the right. I'll restate Equation (1): as we choose values of $$Δx$$ that get closer and closer to zero, what value (the thing that goes on the right-hand side of Equation (1)) does $$x^2$$ get closer and closer to? Hopefully you can see the pattern in Table 1. As we choose values of $$Δx$$ that get closer and closer to zero, $$x^2$$ is getting closer and closer to being $$4$$. Thus, we've found the answer to the problem and we can write

$$\lim_{Δx→0}x^2=4\tag{2}$$

Figure 2: Graph of the function $$f(x)=x^2$$. As the values of $$x$$ get closer and closer to equaling $$2$$, the values of $$x^2$$ get closer and closer to equaling $$4$$.

Using a limit to define the derivative

"Introduction to the idea of a derivative as instantaneous rate of change or the slope of the tangent line."$$^{}$$

This problem might have seemed trivial, but the importance of the notion of a limit is that it can be applied to any$$^1$$ arbitrary function $$f(x)$$ (not just $$x^2$$) and we can let $$Δx$$ approach anything (not just zero). One very important function from algebra to apply the limit to is the function $$\frac{f(x+ Δx)-f(x)}{(x+Δx)-x}$$ which we'll just call $$h(x)$$ to make life easier. Let's take the limit of this function as $$Δx→0$$ to get

$$\lim_{Δx→0}h(x)=?\tag{3}$$

To figure out what $$h(x)$$ gets closer and closer to as $$Δx$$ approaches zero, we'll do the same steps that we did to solve Equation (1). Let's start off with $$Δx=Δx_1$$; then we have $$h(x)=\frac{f(x+ Δx_1)-f(x)}{(x+Δx_1)-x}$$. As you can see from looking at Figure 3, $$h(x)=\frac{f(x+ Δx_1)-f(x)}{(x+Δx_1)-x}$$  gives us the slope of the line $$AB_1$$. Let's choose a smaller value of $$Δx$$. Let's choose $$Δx=Δx_3$$. Then, $$h(x)=\frac{f(x+ Δx_3)-f(x)}{(x+Δx_3)-x}$$. You can see from Figure 3 that this gives the slope of the line $$AB_3$$. If we choose an even smaller value of $$Δx$$ (say $$Δx=Δx_4$$), then $$h(x)=\frac{f(x+ Δx_4)-f(x)}{(x+Δx_4)-x}$$ which gives us the slope of the brown line in Figure 3. The question is: as $$Δx$$ approaches zero, what does the function $$h(x)$$ get closer and closer to equaling? Well, you can see that as $$Δx→0$$, the function $$h(x)$$ gives us the slopes of lines which resemble more and more closely the dashed grey line in Figure 3. We'll give this slope a name and call the slope of the dashed grey line the derivative of $$f(x)$$ at the point $$x)$$. We write the derivative of $$f(x)$$ at $$x$$ like this: $$f'(x)$$ or $$\frac{df}{dx}$$.

To solve the problem we initially started with, we can see that as the values of $$Δx$$ get closer and closer to zero, the values of $$h(x)$$ are getting closer and closer to the slope of the dashed grey line (which we called $$f'(x)$$). Thus, the solution to Equation (3) is

$$\lim_{Δx→0}\frac{f(x+ Δx)-f(x)}{(x+Δx)-x}=f'(x).\tag{3}$$

Equation (3) is the definition of the derivative function.

Figure 3 (click to enlarge): Graph of arbitrary function $$f(x)$$. If we let $$h(x)=\frac{f(x+ Δx)-f(x)}{(x+Δx)-x}$$, then the function $$h(x)$$ gives us the slope of the line obtained by drawing a straight line through the two points $$(x,f(x))$$ and $$(x+Δx,f(x+Δx))$$. For example if $$Δx=Δx_1$$, then $$h(x)=\frac{f(x+ Δx_1)-f(x)}{(x+Δx_1)-x}$$ gives the slope of the line $$AB_1$$ (dark green line) in the graph above. As the value of $$Δx$$ gets closer and closer to equaling zero, the value of $$h(x)$$ gets closer and closer to equaling the slope of the grey line in the graph above.

1. $$g(x)$$ can be any arbitrary function so long that it is smooth and continuous.