# The definition of limit

1. Sep 21, 2007

### flora

What is the definition of limit? i read some documents but it was not clear. Could you explain what epsilon and delta is ?

2. Sep 21, 2007

### fantispug

There are a few different, equivalent, ways of defining the limit of a real function. The first way taught is usually the epsilon-delta definition:

First we need to define a limit point. A limit point $$c$$ of a set $$A$$ is a point such that for any natural number n, the interval $$(c-1\n,c+1\n)$$ contains something in $$A$$ other than c. So If A=[0,1]u{2} then any number in [0,1] is a limit point of A, but 2 is not (because (2-1/2,2+1/2)=(3/2,5/2) doesn't contain anything in A other than 2).
If you don't get limit points don't worry about it for now, it's a bit of a technical thing and you can forget about it for the second, but we need it to give a precise definition of the limit (otherwise funny things can happen).

Let $$x_0$$ be a limit point of the domain of the real function $$f$$. Then $$\lim_{x \rightarrow x_0}f(x)=L$$ if and only if: for every $$\epsilon>0$$ there exists a $$\delta>0$$ such that if
$$\left|a-x_0\right|<\delta|$$ and $$x\neqx_0$$ then $$\left|f(a)-L\right|<\epsilon$$.

Sound straight forward? No not really.
Intuitively speaking the idea is that if we look really close to the point $$x_0$$ we get really close to the limit $$L$$. Epsilon tells you how close you want the limit L to be to the function f(x) around the point x0. Delta tells you how close you need x to be to x0 so that f(x) is within epsilon of L.
Note that the actual behaviour of the function at x0 is irrelevant.

So a brief example: Show $$\lim_{x\rightarrow 2}x^2=4$$
So lets say we've got some epsilon greater than zero. The question is then how close to 2 do we need to be for x^2 to be within epsilon of 4?
Well let us have a look at:
$$\left|f(a)-L\right|=\left|a^2-4\right|<\epsilon$$
What do we need for this to be true? Let's factorise the equation:
$$\left|a-2\right|\left|a+2\right|<\epsilon$$

So to get a quick idea, suppose $$\epsilon=1$$, then we want to find $$a$$ such that
$$\left|a-2\right|\left|a+2\right|<1$$
Well what if a was within .1 of 2 (that is delta=.1)? So $$1.9\leq a\leq 2.1$$ Is that close enough?
Well if that's true $$3.9\leq\left|a+2\right|\leq4.2$$ and $$\left|a-2\right|\leq.1$$
So putting it all together $$\left|a^2-4\right|\leq 4.2*0.1=.42 < 1$$

So it IS good enough (and so any number smaller than .1 would be good enough too). But what if we wanted epsilon=0.1? Or epsilon=.00000095? Or epsilon=2*10^-54? We could keep guessing but it would get time consuming - we want a prescription that automatically produces a delta for a given epsilon.

So what to we need $$\delta$$ to be such that if $$\left|a-2\right|<\delta$$ then
$$\left|a-2\right|\left|a+2\right|=\delta\left|a+2\right|<\epsilon$$?
We need to do something about that a+2 term. Let's say delta is less than 1, ideally we want to make it small, so we'll start here. Then
$$\left|a-2\right|<1$$ implies $$.9\leq a \leq 1.1$$ so $$2.9\leq\left|a+2\right|\leq3.1$$.
Then if $$\delta\left|a+2\right|<\epsilon$$
$$\delta<\frac{\epsilon}{\left|a+2\right|}\leq\frac{\epsilon}{2.9}$$

So if $$\frac{\epsilon}<1$$ then $$\delta=\frac{\epsilon}{3}<\frac{\epsilon}{2.9}$$ ensures that $$\left|a^2-4\right|<\epsilon$$ providing $$\left|a-2\right|<\delta$$.
If $$\epsilon\geq1$$ then you can see that $$\delta=1$$ is good enough. So we take $$\delta=\min(\frac{\epsilon}{3},1)$$.

Finally just to check that it works, look at epsilon=1 again. Our prescription tells us delta=1/3 should work. Then $$\left|a-2|<1/3$$ and $$\left|a+2|<2+1\3=5/3$$ hence $$\left|a^2-4\right|<5/9<1$$, so it works in this case.
So if you tell me you want to be within epsilon of 4, I can tell you that if you're within delta of 2, then x^2 is within epsilon of 4.

Finally note that since we ignore the behaviour of the function at x0 if we had chosen, $$f(x)=x^2$$ for $$x\neq2$$ and $$f(2)=1000000$$ we'd get the same result.

3. Sep 21, 2007

### SiddharthM

look at the definition of a limit of a sequence first.

take X(n) with limit x.
the epsilon here is is basically a 'error' so to speak. It says for any small-error, we can find a point in the sequence, after which, all points in the sequence are within that error of x (think of absolutevalue(x-y) as the distance between x and y, and that distance after a certain point in the sequence is ALWAYS less than epsilon).
Hope that helps, stewart's calculus book has a great illustration of this definition in his chapter on sequences and series. It's not easy so don't worry if you have to think about it for a while.

The definition for a limit of a function is much easier to understand once you have this down.

4. Sep 22, 2007

### mathwonk

the limit of a function f at a, is the number such that redefining f(a) as that number, makes f continuous at a.

5. Sep 22, 2007

### HallsofIvy

Staff Emeritus
Great! And the defiition of continous is?

6. Sep 22, 2007

### haushofer

You define that with those delta's and epsilons: if x gets arbitrarily close to a, then f(x) gets arbitrarily close to f(a). So:

|x-a|<delta should mean that |f(x)-f(a)|<epsilon.

Like said, you demand that if you restrict x to a, then f(x) is restricted to f(a) ( so, if x is very close to a, then f(x) is very close to f(a) and can't make any funny jumps ) Ofcourse, epsilon is a function of delta, otherwise one couldn't make such a restriction. If you can manage to show this, then you say that f(x) is continuous at x=a.

Maybe it helps to draw a discontinuous function in a graph, and see that with x arbitrarily close to a, you can't get f(x) arbitrarily close to f(a), and so you can't make such a restriction.

7. Sep 22, 2007

### flora

stewart's calculus book... which edition should i have

8. Sep 22, 2007

### matt grime

None, unless you have to owing a course you're taking.

9. Sep 22, 2007

### mathwonk

as halls well knows, f is continuous at a if given any open interval J containing f(a), there is some open interval I centered at a, such that for every x in I, if f is defined at x, then f(x) is in J.

but my opinion is that continuity is a more fundamental notion than limit. e.g. in most books, like stewart certainly, the definition of limit is esentially a sham, since it is not used afterwards to do anything.

the essential point is that two functions which agree away from a point a, at least in some interval centered at a, have the same limit at a.

so in stewart and others of that ilk, the limit of a function like (x^2-4)/(x-2) is computed by canceling a factor of (x-2), and then substituting x = 2 in the remaining factor x+2.

this is merely an invocation of my principle, since x+2 is continuous at 2.
the only case where a more complicated argument is used is in the "proof" that the klimit of sin(x)/x is 1 as x-->0.

here the argument uses the squeeze principle (applied to areas): i.e. if a function's graph lies between the graph of two continuous functions which agre at a, at least for x near a and different from a,. then the original function has limit at a equal to the value of either continuous function there.

these applications of continuoius functions are what are actually used in all standard texts, and hence should at least be stated by them.

Last edited: Sep 22, 2007