Limit (mathematics)

<math>\frac{d}{dx} \sin x=?\,</math>

This article/section deals with mathematical concepts appropriate for late high school or early college.

The concept of the limit is the cornerstone of calculus, analysis, and topology. At the simplest intuitive level, the limit of a function at a point is the value that the function "approaches" as its argument "approaches" that point. But that is an unsatisfactory definition, and a much more careful definition is required.

Limit of a Sequence

Let <math>(a_n)_{n\in N} = (a_1, a_2, a_3, ...,</math> be a sequence of real numbers. We say that this sequence has a limit <math>a\,</math>, i.e., <math>\lim_{n\to \infty} = a,\,</math> if for any <math>\varepsilon > 0\,</math>, there exists a number <math>N\,</math>, such that <math> |a_n - a| < \varepsilon\,</math> for every <math>n > N\,</math>. Intuitively, this means that if you take an interval as small as you like, centered at the limit point, then most - i.e., all but a finite number - of the points of the sequence are in that interval.

Limit of a Function

Here is an example. Suppose a function is

<math>f(x) = \frac{x^2-x-6}{x^2-2x-3}\,</math>

What is the limit of f(x) as x approaches 3? This could be written

<math>\lim_{x\to 3}f(x)\,</math>

We can't just evaluate f(3), because the numerator and denominator are both zero. But there's a trick here—we can divide (x-3) into both numerator and denominator, getting

<math>f(x) = \frac{x+2}{x+1}\,</math>

so the limit is 5/4. (That does't actually prove that the limit is 5/4. Once we have defined the limit correctly, we will need a few theorems about limits and continuous functions to establish this result. It is nevertheless true.)

Now try an example that isn't trivial. Let

<math>f(x) = x^x\,</math>

This function is well-defined for x>0, using the general definition that involves the exponential and natural logarithm functions. But we can calculate f(x) for other values:

f(5) = 3125
f(2) = 4
f(0.5) = 0.7071
f(0.3) = 0.6968
f(0.2) = 0.7248

It got smaller, but now it's getting bigger. What's happening?

f(0.1) = 0.7943
f(0.01) = 0.95499
f(0.0001) = 0.999079
f(0.000001) = 0.99998618

It looks as though it's approaching 1. Is it? And what does that mean?

f(1 trillionth) = 0.999999999972369

Remember that f(0) doesn't exist. So we really have to be careful.

We are going to say that the limit of f(x), as x approaches 0, is 1. What that means is this:

We can get f(x) arbitrarily close to 1 if we choose an x sufficiently close to zero. If we want f(x) within one quadrillionth of 1, <math>x = 10^{-17}\,</math> will do. We never have to set x to zero exactly, and we never have to get f(x) = 1 exactly.

In the general case, for arbitrary functions, we might want to say something like

<math>\lim_{x\to X}f(x) = Y\,</math>

For Every Epsilon ...

Stated precisely, given any tolerance ε (by tradition, the letter ε is always used) x being sufficiently close to X will get f(x) within ε of Y. That condition is formally written:

<math>|f(x) - Y| < \varepsilon\,</math>

In this example (<math>f(x) = x^x\,</math>, X=0, Y=1), when ε is <math>10^{-15}\,</math> (one quadrillionth), <math>x < 10^{-17}\,</math> satisfies the condition.

... There Exists a Delta

The way we formalize the notion of x being very close to X is:

"There is a number δ (by tradition it's always δ) such that, whenever x is within δ of X, f(x) is within ε of Y".

That is written:

Whenever <math>0 < |x - X| < \delta, |f(x) - Y| < \varepsilon\,</math>

In our example, if ε is <math>10^{-15}\,</math>, δ = <math>10^{-17}\,</math> works. That is, any <math>x < 10^{-17}\,</math> will guarantee <math>|f(x)-1| < 10^{-15}\,</math>.

The Full Definition

So here is the full definition:

<math>\lim_{x\to X} f(x) = Y\,</math> means

For every ε > 0, there is a δ > 0 such that, whenever <math>0 < |x-X| < \delta, |f(x)-Y| < \varepsilon\,</math>

So the definition is sort of like a bet or a contract—"For any epsilon you can give me, I can come up with a delta."

A few things to note:

• We require ε > 0. Specifying a required tolerance of zero is not allowed. We only have to be able to get f(x) within an arbitrarily close but nonzero tolerance of Y. We don't ever have to get it exactly equal to Y.
• We have 0 < |x-X| < δ, not just |x-X| < δ. That is, we never have to calculate f(X) exactly. f(X) doesn't need to be defined. In the example we are considering, <math>0^0\,</math> isn't defined.

This definition, and variations of it, are the central point of calculus, analysis, and topology. The phrase "For every ε there is a δ" is ingrained into the consciousness of every mathematics student. This notion of a "bet" could be considered to set the branches of mathematics that follow (calculus, topology, ...) apart from the earlier branches (arithmetic, algebra, geometry, ...) Students who have mastered the notion of "For every ε there is a δ" are ready for higher mathematics.

In our example of <math>f(x) = x^x\,</math>, we haven't actually satisfied the definition of the limit, because f(x) isn't defined for negative x. There are more restrictive notions of "limit from the left" and "limit from the right". We have the limit from the right of <math>x^x = 1\,</math>, which means

For every ε > 0, there is a δ > 0 such that, whenever 0 < x-X < δ, |f(x)-Y| < ε

We still haven't proved that the limit is actually 1. We just gave some accurate calculations strongly suggesting that it is. In fact it is, and the proof requires a few theorems about limits, continuous functions, and the exponential and natural logarithm functions.

The Limit of f(x) is Infinity

With this precise (or, as mathematicians say, rigorous) definition of a limit, we can examine variations that involve "infinity". Remember, Infinity is not a number. It is only through the magic of the epsilon-delta formulation that we can make sense of it.

We might say something like "The limit of f(x), as x approaches X, is infinity." What that means is that we replace the

<math>|f(x) - Y| < \varepsilon\,</math>

with

<math>f(x) > M\,</math>

That is,

For every M > 0, there is a δ > 0 such that, whenever <math>0 < |x-X| < \delta, f(x) > M\,</math>

Consider the function

<math>f(x) = \frac{1}{(x-3)^2}\,</math>

If you graph this, there is an "infinitely high" peak at x=3. We have:

<math>\lim_{x\to 3}f(x) = \infty\,</math>

If we set <math>M = 10^{20}\,</math>, <math>\delta = 10^{-10}\,</math> will win the bet.

For a limit of minus infinity, we set M to some huge negative number, so

<math>\lim_{x\to X}f(x) = - \infty\,</math>

means

For every M < 0, there is a δ > 0 such that, whenever <math>0 < |x-X| < \delta, f(x) < ,</math>