A question about the epsilon delta definition of a limit

[TooManyHours]

Hi, I have a question about the epsilon / delta definition of limits, for example the limit of x as it approaches c for f(c) = L.

As I understand it, epsilon is basically the number of units on either side of L on the y-axis that makes a range between L + epsilon and L – epsilon with L being in the exact middle of the range.

There is supposed to be a corresponding range on the x-axis around c using delta units with the range between c + delta and c – delta with c being in the exact middle of the range.

And according to the definition, as long as you pick some x value that falls between c + delta and c – delta, the f(x) function will always produce a value that falls between L + epsilon and L – epsilon range. Sounds straight forward enough.

However, doesn’t this imply that if f(c) = L, then f(c + delta) should be L + epsilon and f(c - delta) should be L – epsilon. That's what all the diagrams in the textbooks show. They show graphs with symmetric ranges.

However, a simple example shows that this is not true.

Take the limit of x^2 as x approaches 5 for example. It’s Limit value is 25.

If I choose a delta of 1 unit, my x-axis range goes from 16 [(5-1)^2] to 36 [(5+1)^2).

If I take the absolute value of f(x) – L for x = 4 then 16 – 25 makes an epsilon of 9 units from L.
If I take the absolute value of f(x) – L for x = 6 then 36 – 25 makes an epsilon of 11 units from L.

Clearly the limit is not in the exact middle of this range or epsilon can have two different values.

Does this mean that this epsilon delta definition only works for delta values that are very very close to c, for example you have to pick a delta of .000001 or even smaller to get values that work out a symmetrical range for delta and epsilon?

Thanks

 

[JThompson]

If I choose a delta of 1 unit, my x-axis range goes from 16 [(5-1)^2] to 36 [(5+1)^2)

This is the incorrect order for delta-epsilon proofs. You don't choose a delta value and find an epsilon. The definition says nothing about being able to find a particular epsilon based on any arbitrarily chosen delta value.

Does this mean that this epsilon delta definition only works for delta values that are very very close to c, for example you have to pick a delta of .000001 or even smaller to get values that work out a symmetrical range for delta and epsilon?

Not necessarily. Suppose I want |x^2-25|<50. Clearly x^2 has a minimum at 0, but the definition doesn't say that f(x) has to take on the value L-epsilon- it just has to be greater than it. So
|x^2-25|<50
-25<x^2<75 we know that x^2 is always positive, so replaced -25 with 0
0<x<sqrt(75)
-5<x-5<sqrt(75)-5
Choose delta to be min{-5, sqrt(75)-5}=sqrt(75)-5. And then of course, reverse the process for the formal proof.
There is no x for which x^2=-50, but there doesn't have to be delta-epsilon doesn't require anything of the sort.EDIT: I just noticed that you wrote "delta values that are very very close to c".. I thought it said "epsilon values that are very very close to L". Delta values don't have to be close to c- just remember, you don't choose delta values arbitrarily.. When you choose a delta, it's because you've already figured out that it will satisfy an epsilon.

 

[TooManyHours]

This is the incorrect order for delta-epsilon proofs. You don't choose a delta value and find an epsilon. The definition says nothing about being able to find a particular epsilon based on any arbitrarily chosen delta value.

The problem I have is that the textbook diagrams seem to imply that for f(c) = L, epsilon is a single value (the number of units above and below L) and delta is a single value (the number of units right and left of c). The diagrams seem to imply that f(c+delta) = L + epsilon and f(c-delta) = L - epsilon.

However, if the correct order is to pick an epsilon value and then figure our which deltas match L + epsilon and L - epsilon, I would still end up with the same problem in that my deltas will not produce a symmetrical range around c.

In my example for x^2 with c = 5 and a limit of 25, if I pick an epsilon = 1 unit away from L, then the epsilon range above and below L is L + epsilon = 26 and L - epsilon = 24.

But the absolute value of [sqrt(26) - 5] = .099 is not the same as the absolute value of [sqrt(24) - 5] = .101

Then what is the range on the x-axis for c + delta and c - delta? Is delta .101 or is it .099? It can't be both.

Then how do you get f(c+delta) to = L + epsilon and f(c-delta) to = L - epsilon for a single delta value and a single epsilon value?

 

[deluks917]

^The textbook is giving you the completely wrong idea. You are right for many functions this "symmetric" range cannot be set up. But that's not the idea. Look at the definition. You need to make |f(x) - L| small for all x on some interval around a.

lim x ->a = L just means if x is close to a then f(x) is close to L. Epsilon and delta are just there to formalize this idea. The idea is that if I give you any epsilon greater then zero you can find a interval around a such that if x is in that interval but x!=a (0< |x-a| < delta)| then |f(x) - L|< epsilon.

You don't have to choose the largest possible delta. Any delta that "gets the job done" will do. For example take x^2. Let's prove lim x->a x^2 = a ^2. We need a relationship between how small |x-a| is and how small |x^2 - a ^2| is. Let's find one. |x^2 - a^2| = |x-a|*|x+a|. This is an encouraging result. Let's make |x-a| < 1 this means:

|x| < |a| + 1. Then |x^2 - a ^2| < |x-a| * |x + a| <|x-a| * (|x| +|a|) <|x-a| * (2|a| + 1).

This gives us what we need. If |x-a| < epsilon/(2|a| +1) then |x^2 - a^2| < |x-a| * (2|a| + 1)< [epsilon/(2|a| +1)]*(2|a| + 1) = epsilon.
Note also that a functions value at a does not change its limit at a as lim x->a does not have to be equal to f(a) if that is true f(x) is continuous at a. Hence we have proved x^2 is continuous at every real number a.

Note we didn't necessarily find the largest epsilon possible. Let's test this works. Say we want |x^2 - 5^2| < .01 (Pretty close). We proved we need |x-a| < .01/11. that means (aproximately) 4.9991<x<5.00091. x^2 is increasing on this interval so we need only check that the endpoints are in (24.99, 25.01). (Note the endpoints aren't technically included but if the endpoints work so will all the interior points).(5-.01/11) ^2 = 24.991, (5 + .01/11) ^2 = 24.0091. This is very close to tolerance but note we do not have exact equality.

Maybe you need to see a "real" application of the limit. When I was trying to learn from Rogawski (a calculus text) I got the wrong idea because I never saw the "formal definition of a limit" used for anything. Try to understand the proof that if f is continuous at a and g is cont at f(a) the g(f(x)) is continuous at a. Intuitively this should be true. Very similar proofs to this one appear throughout analysis.

---------------
Assumptions: (the letters are changed to emphasize they are not the same and that that the names don't matter)

1) g is cont at f(a). This means lim y -> a g(y) = g(f(a)). Restated again it says for every epsilon greater then zero there exists a delta greater then zero s.t. if |y-f(a)| < delta then |g(y) - g(f(a))| < epsilon. (why don't we write 0<|y-g(f(a))|<delta?)

2) f is cont at a. Or for every theta>0 there exists a zeta greater then zero s.t. if |x-a| < zeta then |f(x) - f(a)| < theta.What we need to show:

for every gamma>0 there exists a lambda > 0 s.t. if |x-a| < lambda then |g(f(x)) - g(f(a))| < gamma.

----------

This is not hard to show once we write everything down! By assumption 1 for every gamma>0 there exists a theta>0 s.t. if |y-f(a)| < theta then |g(y) - g(f(a))| < gamma. But by assumption 2 we can find a lambda>0 s.t. if |x-a| <lambda then |f(x) - f(a)| < theta (we can find this lambda for any zeta even theta!). This means if |x-a|< lambda then |f(x)-f(a)| < theta which means |g(f(x))- g(f(a))| < gamma. That's the proof.

 

[TooManyHours]

^The textbook is giving you the completely wrong idea. You are right for many functions this "symmetric" range cannot be set up. But that's not the idea. Look at the definition. You need to make |f(x) - L| small for all x on some interval around a.

lim x ->a = L just means if x is close to a then f(x) is close to L. Epsilon and delta are just there to formalize this idea. The idea is that if I give you any epsilon greater then zero you can find a interval around a such that if x is in that interval but x!=a (0< |x-a| < delta)| then |f(x) - L|< epsilon.

Thanks, I think I'm getting it now.

Based on your previous example, I wrote up my own even simpler example:

Hopefully this is correct:

The limit as x approaches 2 of 2x = 4.

For epsilon > 0 we need to find a delta > 0 such that 0 < |x - 2| < delta then |2x - 4| < epsilon.

So we need a connection between |x - 2| and |2x - 4|.

The connection is that |2x - 4| = 2|x - 2|.

To make |2x - 4| < epsilon we need to make 2|x - 2| < epsilon.

We do this by making |x - 2| < epsilon/2 and therefore a delta = epsilon/2.

To show that this delta will work, we can show that...

if 0 < |x - 2| < epsilon/2 then 2|x - 2| < epsilon.

And by our connection above, 2|x - 2| = |2x - 4| and therefore |2x - 4| < epsilon.

Hope that's it...

The difficulty comes from the fact that in high school, limits are shown simply as a different type of function to evaluate without any real defintion other than some pictures showing some value getting closer and closer to some other value as a basis for finding a slope at some point on curve (i.e. a derivative). You never had to prove anything, just calculate stuff.

When a prof in universty writes the formal epsilon delta definition on a blackboard using something called the first order predicate caluclus, you swear you can hear heads exploding in the room. It looks like he wrote something in Klingon.

Alas, it appears that once you get to university, there's no numbers in the math any more. In other words, my fancy calculator is now kind of useless.

 

[deluks917]

Yes that is the correct proof. If things seem confusing in math class just remember in math that many concepts that are easy to intuitively understand can be difficult to formalize. For example we all know f(x) = cx is continuous but to prove it (do you see how to prove it?) requires the above definition of a limit (or some equally abstract definition). In the end these definitions are useful when we handle a function that is not so easy to intuitively understand. In math I try to focus on why we defined things the way we did and why the theorems are important.

 

[TooManyHours]

Yes that is the correct proof. If things seem confusing in math class just remember in math that many concepts that are easy to intuitively understand can be difficult to formalize. For example we all know f(x) = cx is continuous but to prove it (do you see how to prove it?) requires the above definition of a limit (or some equally abstract definition). In the end these definitions are useful when we handle a function that is not so easy to intuitively understand. In math I try to focus on why we defined things the way we did and why the theorems are important.

Thanks for the help.

In high school, math was taught as being a collection of tools to use to solve some problem. In universtiy I guess they figure we know how to use some of those tools, now we need to understand how some of those tools were made and why they actually work.