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Line 13: − Let <math>f(x)</math> be a real valued [[function]] in one [[variable]]. We say that+ + + + + + + + − + − \lim_{x\to p} f(x) = L+ − + + + + + + + + + + + + + − if for any error <math>\varepsilon</math>, we can find a sufficiently small [[neighborhood]] of <math>p</math> so that <math>f(x)</math> is within <math>\varepsilon</math> of the value <math>L</math> for any <math>x\neq p</math> in that neighborhood. One says that the limit of ''f(x)'' at ''p'' exists and is equal to ''L''.+ − The standard, though more verbose, way of saying this is that for any <math>\varepsilon > 0</math> there exists a sufficiently small <math>\delta > 0</math> such that <math>|f(x)-L| <\varepsilon</math> whenever <math>0<|x-p|<\delta</math>.+ − Intuitively, this means that as the variable <math>x</math> approaches the value <math>p</math>, the function <math>f(x)</math> tends to the value <math>L</math>.+ + + + + + + + + + + + + + + + + + + + + + + + +
→Limit of a Function: Major rewrite. I think this is accessible and rigorous for the target group.
==Limit of a Function==
==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 function]]s to establish this result. It is nevertheless true.)
<math>
Now try an example that isn't trivial. Let
:<math>f(x) = x^x</math>
</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>
"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.
[[category:mathematics]]
[[category:mathematics]]