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. As we will see, 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 involvels 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 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. We never have to set x to zero exactly. If we want x within one quadrillionth of 1, <math>x = 10^{-17}</math> will do.

Stated precisely, given any tolerance ε (by tradition, the letter ε is always used) x being sufficiently close to 0 will get f(x) within ε of 1. That condition is formally written:

<math>|f(x)-1| < \varepsilon</math>

In this example, when ε is one quadrillionth, <math>x < 10^{-17}</math> satisfies the condition.

"x is sufficiently close to zero" is formalized as

"There is a number δ (by tradition it's always δ) such that, whenever x is within δ of 0, f(x) is within ε of 1".

That is written:

Whenever <math>0 < |x-0| < \delta, |f(x)-1| < \varepsilon</math>

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>

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 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 conciousness of every mathematics student.

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 function]]s, and the exponential and natural logarithm functions.