# 1.1: Drawing Tangents and a First Limit

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Now — our treatment of limits is not going to be completely mathematically rigorous, so we won't have too many formal definitions. There will be a few mathematically precise definitions and theorems as we go, but we'll make sure there is plenty of explanation around them.

Let us start with the “tangent line” problem. Of course, we need to define “tangent”, but we won't do this formally. Instead let us draw some pictures.

Here we have drawn two very rough sketches of the curve $$y=x^2$$ for $$x \geq 0\text{.}$$ These are not very good sketches for a couple of reasons

• The curve in the figure does not pass through $$(0,0)\text{,}$$ even though $$(0,0)$$ lies on $$y=x^2\text{.}$$
• The top-right end of the curve doubles back on itself and so fails the vertical line test that all functions must satisfy 1 — for each $$x$$-value there is exactly one $$y$$-value for which $$(x,y)$$ lies on the curve $$y=x^2\text{.}$$

So let's draw those more carefully.Figure 1.1.1 Sketches of the curve $${y=x^2}\text{.}$$ (left) shows a tangent line, while (right) shows a line that is not a tangent.

These are better. In both cases we have drawn $$y=x^2$$ (carefully) and then picked a point on the curve — call it $$P\text{.}$$ Let us zoom in on the “good” example:

We see that as we zoom in on the point $$P\text{,}$$ the graph of the function looks more and more like a straight line. If we kept on zooming in on $$P$$ then the graph of the function would be indistinguishable from a straight line. That line is the tangent line (which we have drawn in blue). A little more precisely, the blue line is “the tangent line to the function at $$P$$”. We have to be a little careful, because if we zoom in at a different point, then we will find a different tangent line.

Now let's zoom in on the “bad” example we see that the blue line looks very different from the function; because of this, the blue line is not the tangent line at $$P\text{.}$$

Here are a couple more examples of tangent lines

The one on the left is very similar to the good example on $$y=x^2$$ that we saw above, while the one on the right is different — it looks a little like the “bad” example, in that it crosses our function the curve at some distant point. Why is the line in Figure 1.1.4(right) a tangent while the line in Figure 1.1.1(right) not a tangent? To see why, we should again zoom in close to the point where we are trying to draw the tangent.

As we saw above in Figure 1.1.3, when we zoom in around our example of “not a tangent line” we see that the straight line looks very different from the curve at the “point of tangency” — i.e. where we are trying to draw the tangent. The line drawn in Figure 1.1.4(right) looks more and more like the function as we zoom in.

This example raises an important point — when we are trying to draw a tangent line, we don't care what the function does a long way from the point; the tangent line to the curve at a particular point $$P\text{,}$$ depends only on what the function looks like close to that point $$P\text{.}$$

To illustrate this consider the sketch of the function $$y = \sin(x)$$ and its tangent line at $$(x,y)=(0,0)\text{:}$$

As we zoom in, the graph of $$\sin(x)$$ looks more and more like a straight line — in fact it looks more and more like the line $$y=x\text{.}$$ We have also sketched this tangent line. What makes this example a little odd is that the tangent line crosses the function. In the examples above, our tangent lines just “kissed” the curve and did not cross it (or at least did not cross it nearby).

Using this idea of zooming in at a particular point, drawing a tangent line is not too hard. However, finding the equation of the tangent line presents us with a few challenges. Rather than leaping into the general theory, let us do a specific example. Let us find the the equation of the tangent line to the curve $$y=x^2$$ at the point $$P$$ with coordinates 2 $$(x,y)=(1,1)\text{.}$$

To find the equation of a line we either need

• the slope of the line and a point on the line, or
• two points on the line, from which we can compute the slope via the formula

$m = \frac{y_2 - y_1}{x_2 - x_1} \nonumber$

and then write down the equation for the line via a formula such as

$y = m \cdot(x - x_1) + y_1. \nonumber$

We cannot use the first method because we do not know what the slope of the tangent line should be. To work out the slope we need calculus — so we'll be able to use this method once we get to the next chapter on “differentiation”.

It is not immediately obvious how we can use the second method, since we only have one point on the curve, namely $$(1,1)\text{.}$$ However we can use it to “sneak up” on the answer. Let's approximate the tangent line, by drawing a line that passes through $$(1,1)$$ and some nearby point — call it $$Q\text{.}$$ Here is our recipe:

• We are given the point $$P=(1,1)$$ and we are told

Find the tangent line to the curve $$y=x^2$$ that passes through $$P = (1,1)\text{.}$$

• We don't quite know how to find a line given just 1 point, however we do know how to find a line passing through 2 points. So pick another point on the curves whose coordinates are very close to $$P\text{.}$$ Now rather than picking some actual numbers, I am going to write our second point as $$Q = (1+h, (1+h)^2)\text{.}$$ That is, a point $$Q$$ whose $$x$$-coordinate is equal to that of $$P$$ plus a little bit — where the little bit is some small number $$h\text{.}$$ And since this point lies on the curve $$y=x^2\text{,}$$ and $$Q$$'s x-coordinate is $$1+h\text{,}$$ $$Q$$'s y-coordinate must be $$(1+h)^2\text{.}$$

If having $$h$$ as an variable rather than a number bothers you, start by thinking of $$h$$ as $$0.1\text{.}$$

• A picture of the situation will help.
• This line that passes through the curve in two places $$P$$ and $$Q$$ is called a “secant line”.
• The slope of the line is then

\begin{align*} m &= \frac{y_2 - y_1}{x_2-x_1}\\ &= \frac{(1+h)^2-1}{(1+h)-1} = \frac{1+2h+h^2-1}{h} = \frac{2h+h^2}{h} = 2+h \end{align*}

where we have expanded $$(1+h)^2 = 1+2h+h^2$$ and then cleaned up a bit.

Now this isn't our tangent line because it passes through 2 nearby points on the curve — however it is a reasonable approximation of it. Now we can make that approximation better and so “sneak up” on the tangent line by considering what happens when we move this point $$Q$$ closer and closer to $$P\text{.}$$ i.e. make the number $$h$$ closer and closer to zero.

First look at the picture. The original choice of $$Q$$ is on the left, while on the right we have drawn what happens if we choose $$h'$$ to be some number a little smaller than $$h\text{,}$$ so that our point $$Q$$ becomes a new point $$Q'$$ that is a little closer to $$P\text{.}$$ The new approximation is better than the first.

So as we make $$h$$ smaller and smaller, we bring $$Q$$ closer and closer to $$P\text{,}$$ and make our secant line a better and better approximation of the tangent line. We can observe what happens to the slope of the line as we make $$h$$ smaller by plugging some numbers into our formula $$m=2+h\text{:}$$

\begin{align*} h=0.1 && m = 2.1\\ h=0.01 && m= 2.01\\ h=0.001 && m= 2.001. \end{align*}

So again we see that as this difference in $$x$$ becomes smaller and smaller, the slope appears to be getting closer and closer to $$2\text{.}$$ We can write this more mathematically as

\begin{align*} \lim_{h \to 0} \frac{(1+h)^2-1}{h} &= 2 \end{align*}

The limit, as $$h$$ approaches $$0\text{,}$$ of $$\frac{(1+h)^2-1}{h}$$ is $$2\text{.}$$

This is our first limit! Notice that we can see this a little more clearly with a quick bit of algebra:

\begin{align*} \frac{(1+h)^2-1}{h} &= \frac{(1+2h+h^2)-1}{h}\\ &= \frac{2h+h^2}{h} & (2+h)\\ \end{align*}

So it is not unreasonable to expect that

\begin{align*} \lim_{h \to 0} \frac{(1+h)^2-1}{h} &= \lim_{h \to 0} (2+h) = 2. \end{align*}

Our tangent line can be thought of as the end of this process — namely as we bring $$Q$$ closer and closer to $$P\text{,}$$ the slope of the secant line comes closer and closer to that of the tangent line we want. Since we have worked out what the slope is — that is the limit we saw just above — we now know the slope of the tangent line is $$2\text{.}$$ Given this, we can work out the equation for the tangent line.

• The equation for the line is $$y=mx+c\text{.}$$ We have 2 unknowns $$m$$ and $$c$$ — so we need 2 pieces of information to find them.
• Since the line is tangent to $$P = (1,1)$$ we know the line must pass through $$(1,1)\text{.}$$ From the limit we computed above, we also know that the line has slope $$2\text{.}$$
• Since the slope is $$2$$ we know that $$m=2\text{.}$$ Thus the equation of the line is $$y=2x+c\text{.}$$
• We know that the line passes through $$(1, 1)\text{,}$$ so that $$y=2x+c$$ must be $$1$$ when $$x=1\text{.}$$ So $$1 = 2 \cdot 1 + c\text{,}$$ which forces $$c = -1\text{.}$$

So our tangent line is $$y=2x-1\text{.}$$

## Exercise

##### Exercise $$\PageIndex{1}$$

On the graph below, draw:

1. The tangent line to $$y=f(x)$$ at $$P\text{,}$$
2. the tangent line to $$y=f(x)$$ at $$Q\text{,}$$ and
3. the secant line to $$y=f(x)$$ through $$P$$ and $$Q\text{.}$$

##### Exercise $$\PageIndex{2}$$

Suppose a curve $$y=f(x)$$ has tangent line $$y=2x+3$$ at the point $$x=2\text{.}$$

1. True or False: $$f(2)=7$$
2. True or False: $$f(3)=9$$
##### Exercise $$\PageIndex{3}$$

Let $$L$$ be the tangent line to a curve $$y=f(x)$$ at some point $$P\text{.}$$ How many times will $$L$$ intersect the curve $$y=f(x)\text{?}$$

This page titled 1.1: Drawing Tangents and a First Limit is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform.