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### Formal Definition of a Limit

Date: 08/01/98 at 04:51:29
From: Michael Sabroski
Subject: Formal Definition of a Limit

The formal definition of the limit states:

Let f be a function that is defined on an open interval
containing c, except possibly at c itself, and let L be
a real number. The statement:

lim x->c f(x) = L

means for every epsilon > 0, there exists a delta > 0,
such that:

if 0 < abs(x - c) < delta, then abs(f(x) - L) < epsilon

So what I'm asking is what does all this mean? From my understanding,
first you have to set up the open interval (L-epsilon, L+epsilon) and
the horizontal lines (y = L - epsilon) and (y = L + epsilon). Then you
pick the open interval on the x-axis, (c-delta, c+delta). If there
exists an open interval (c-delta, c+delta) such that for every x in
(c-delta, c+delta), with the possible exception of x = c, the point
P(x,f(x)) lies between the horizontal lines, then:

(L-epsilon) < f(x) < (L+epsilon)

and hence lim x->c f(x) = L.

What I'm confused about is the connection between the open intervals
and the informal definition of the limit? Why even bother with
intervals? And do epsilon and delta get smaller over time? It seems to
me that the lim (delta->0) epsilon = 0. This should be true because x
gets closer to c and f(x) gets closer to L. Or am I just overthinking
it?

Date: 08/05/98 at 14:43:36
From: Doctor Benway
Subject: Re: Formal Definition of a Limit

Overthinking it? Impossible! I've always said that if you walk away
from thinking about a math concept with fewer questions about it than
you had before you started thinking about it, then you didn't think
hard enough. Well, enough of my ranting, I'll try to help you with the
problem.

"For every epsilon there is a delta ..." basically means the following.
Suppose you have some function f(x), and you run across the statement
lim x->5 f(x) = 3. All this means is that you can take a tiny interval
around 3 and find a tiny interval around 5 such that every x in the
little interval around 5 lands in the little interval around 3 when you
take f(x).

For example, say we take a little interval around 3 from 2.999 to
3.001. Then we can find some tiny interval around 5 where everything
in that interval (except maybe 5 itself) will go to some number
inside the little interval around 3. In other words, "if you take a
number really close to 5 and run it through the function, then you
get a number really close to 3."  Where the formal part of the
definition comes in is that you can take any interval around 3, no
matter how tiny, and still come up with some interval around 5 that
will work.

Now to attack your other questions. What is this stuff about open
intervals? In this case, open intervals can be thought of as
"everything that is a distance less than _____ away from something."
Say you want everything that is a distance less than .0001 away from 5.
You will end up with an open interval from 4.9999 to 5.0001, excluding
the endpoints. The endpoints are excluded because the endpoints are
exactly a distance of .0001 away from 5. Ponder this before moving on.
Since you have excluded endpoints, then you have an open interval.
Since our definition of limit involves "all the stuff that's this close
to some number," you can see how open intervals come up.

As for epsilon and delta shrinking over time, well, really they don't
exactly. As I pointed out above, you can make epsilon as small as you
want and still find a delta that works, if you have a limit.

Your last statement about the limit of delta as epsilon goes to zero
being zero is a little dodgy too, just from formal definitions. The
explanation of this is very mathematical, so if you're not feeling up
to it, you can skip the next few sentences.

Limits as we are discussing them here only work with functions, and
delta is not a function of epsilon. Recall that a function has only one
element paired up with everything in its domain. Say you pick some
value for epsilon and end up with a delta that works. Well, anything
smaller than that delta will also work, meaning you have more than one
thing matched up with that epsilon so we don't have a function.

End of formal stuff. What you are saying is basically true, if you take
smaller and smaller epsilons you have to take smaller and smaller
deltas in order to land in the "zone" you want, but technically it's
not quite correct. Math is funny like that. When you get into higher
math you have to be very careful and very formal about things, because
not everything you think *should* work really does work. At least it
keeps things interesting.

Well, hope I at least partially answered some of your questions and
maybe even raised new ones for you. Thanks for writing.

- Doctor Benway, The Math Forum
Check out our web site! http://mathforum.org/dr.math/

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