Limits: An Introduction

Calculus was used long before it was established on firm mathematical foundations. Limits provide a precise way of talking about convergence and infinite processes.

For example, derivatives and integrals are defined using limits. You'll also use limits to study graphs.

Intuitively, convergence means that a variable quantity approaches a fixed number. For example, consider $f(x) = \dfrac{x - 2}{x^2 - 4}$ . Plug in numbers close to 2:

$$\vbox{\offinterlineskip \halign{ & \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & x & & $f(x)$ & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 2.1 & & 0.24390 & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 1.99 & & 0.25063 & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & \cr & 2.001 & & 0.24994 & \cr height2pt & \omit & & \omit & \cr \noalign{\hrule} }} $$

It seems as though the $f(x)$ -values are close to 0.25. If you graph $\dfrac{x - 2}{x^2 - 4}$ , the picture seems to confirm this:

$$\hbox{\epsfysize=2in \epsffile{limintr1.eps}}$$

Observe that $\dfrac{x - 2}{x^2 - 4}$ is not defined at $x = 2$ . In thinking about the limit of a function $f(x)$ as x approaches c, you don't consider what happens when x equals c; you consider what happens when x is close to c.

In this case, when x is close to 2, it appears that $\dfrac{x - 2}{x^2 - 4}$ is close to 0.25. The mathematical expression is: {\it The limit of $\dfrac{x - 2}{x^2 -
   4}$ as x approaches 2 is 0.25.} In symbols,

$$\lim_{x\to 2} \dfrac{x - 2}{x^2 - 4} = 0.25.$$

In general, to say that

$$\lim_{x \to a} f(x) = L$$

means that $f(x)$ can be made arbitrarily close to L for all x's sufficiently close to a.

I'll discuss the definition and some rules for computing limits later. First, I'll show you some computations so you can get a feel for the ideas.


Example. Compute $\displaystyle \lim_{x \to 2} \dfrac{x - 2}{x^2 - 4}$ .

If you plug 2 into $\dfrac{x - 2}{x^2 -
   4}$ , you get $\dfrac{0}{0}$ . This is called an indeterminate form. This means that you can't conclude anything from the form $\dfrac{0}{0}$ : The limit might be a number, it might be infinite, or it might be undefined.

When plugging in yields an indeterminate form, you have to do more work before you can come to a conclusion. "More work" often involves algebraic simplification.

In this case, I fact $x^2 - 4$ , then cancel $x - 2$ 's:

$$\lim_{x \to 2} \dfrac{x - 2}{x^2 - 4} = \lim_{x \to 2} \dfrac{x - 2}{(x - 2)(x + 2)} = \lim_{x \to 2} \dfrac{1}{x + 2} = \dfrac{1}{4}.$$

Why am I allowed to cancel the $x - 2$ 's? I noted earlier that in computing $\displaystyle \lim_{x \to 2}
   \dfrac{x - 2}{x^2 - 4}$ I only consider x's near 2, not x equal to 2. Since $x \ne 2$ , I have $x - 2 \ne 0$ , so cancellation is legal.

I did the last step by plugging $x =
   2$ into $\dfrac{1}{x + 2}$ . This time I did not get an indeterminate form, and the rules for limits I'll discuss later tell me that $\dfrac{1}{4}$ is the answer.


I won't always describe the action in such excruciating detail, but you should understand why the algebraic manipulations are legitimate. They usually reduce to the idea in the last example.


Example. Compute $\displaystyle \lim_{x \to 1} \dfrac{x^3 - 1}{x - 1}$ .

If you plug $x = 1$ into $\dfrac{x^3 - 1}{x
   - 1}$ , you get $\dfrac{0}{0}$ . This means you have more work to do.

Since

$$x^3 - 1 = (x - 1)(x^2 + x + 1),$$

it follows that

$$\lim_{x \to 1} \dfrac{x^3 - 1}{x - 1} = \lim_{x \to 1} \dfrac{(x - 1)(x^2 + x + 1)}{x - 1} = \lim_{x \to 1} (x^2 + x + 1) = 3.\quad\halmos$$


Example. Compute $\displaystyle \lim_{x \to 2} (4x^3 - 5x + 11)$ .

If I plug $x = 2$ into $4x^3 - 5x + 11$ , I get $32 - 10 + 11 = 33$ . This is not an indeterminate form; it's just a number. In fact,

$$\lim_{x \to 2} (4x^3 - 5x + 11) = 33.\quad\halmos$$


The final steps in the last two examples are special cases of the following general rule:

$$\lim_{x\to c} p(x) = p(c).$$

That is, you can compute the limit of a polynomial by "plugging the number in". When you can compute $\displaystyle \lim_{x\to c} f(x)$ by plugging in $x = c$ (to get $f(c)$ ), the function f is continuous at $x
   = c$ . I'll discuss continuity in more detail later.


Example. Compute $\displaystyle \lim_{x \to 2} \dfrac{x - 2}{\dfrac{1}{2} -
   \dfrac{1}{x}}$ .

Plugging in gives $\dfrac{0}{0}$ . I have more work to do. Add the fractions on the top and simplify:

$$\lim_{x \to 2} \dfrac{x - 2}{\dfrac{1}{2} - \dfrac{1}{x}} = \lim_{x \to 2} \dfrac{x - 2}{\dfrac{x - 2}{2x}} = \lim_{x \to 2} \dfrac{2x(x - 2)}{x - 2} = \lim_{x \to 2} 2x = 4.$$

I got the last equality by plugging 2 into $2x$ and using the rule for polynomials. Notice a common thread in the last few problems. If plugging into produces a $\dfrac{0}{0}$ form, something must be producing the 0's. Often it is a common factor, which can be cancelled from the top and bottom when you've identified it.


Example. Compute $\displaystyle \lim_{x \to 3} \dfrac{x - 3}{\sqrt{x} -
   \sqrt{3}}$ .

If you plug $x = 3$ into $\dfrac{x -
   3}{\sqrt{x} - \sqrt{3}}$ , you get $\dfrac{0}{0}$ . This means you have some work to do.

First,

$$x - 3 = (\sqrt{x} - \sqrt{3})(\sqrt{x} + \sqrt{3}).$$

Therefore,

$$\lim_{x \to 3} \dfrac{x - 3}{\sqrt{x} - \sqrt{3}} = \lim_{x \to 3} \dfrac{(\sqrt{x} - \sqrt{3})(\sqrt{x} + \sqrt{3})} {\sqrt{x} - \sqrt{3}} = \lim_{x \to 3} (\sqrt{x} + \sqrt{3}) = 2\sqrt{3}.$$

I didn't get an indeterminate form when I plugged in, so it's reasonable that the last step is valid. This seems to be confirmed by the graph; $2\sqrt{3} \approx 3.5$ .

$$\hbox{\epsfysize=2in \epsffile{limintr2.eps}}\quad\halmos$$


Example. Compute $\displaystyle \lim_{x \to 1} \dfrac{x^2 - 2x - 3}{x - 1}$ .

Plugging in gives $\dfrac{-4}{0}$ . The limit is undefined.

The graph shows a vertical asymptote at $x = 1$ :

$$\hbox{\epsfysize=2in \epsffile{limintr3.eps}}$$

If I plug in values of x near 1, I get a wide range of outputs:

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & x & & 0.9 & & 0.999 & & 1.0003 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & $f(x)$ & & 39.9 & & 4000 & & -13333.3 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

These empirical results seem to confirm that the limit is undefined.


The general rule is:


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