Studying Mathematics Thread

[GrandizerII]

This is something I blogged about a couple years back (yes, it's not 100% accurate/applicable, but it's good enough):

I’ve been reviewing calculus online lately, currently doing limits specifically. Overall, the concept of the limit is easy to grasp intuitively. But mathematically, it can be quite challenging to understand the precise definition of the limit (I’m referring to the epsilon-delta definition of the limit). And I’ve been struggling the whole day trying to really understand what the definition is actually saying. I think I finally get it now, but let me see if I can explain it in writing here on this page.

So the definition goes like this:

Have function f(x) defined on an interval containing x=a, except possibly at x=a. The limit of f(x), as x approaches a, is equal to L means that for every “epsilon” (that is greater than zero), there is a “delta” (also greater than zero), such that |f(x) – L| < “epsilon” whenever zero < |x – a| < “delta”.

Confusing, yeah? Don’t feel bad if you do find this confusing. Apparently, even calculus graduates struggle with really understanding the definition.

Anyway, let’s see how to go about explaining the definition:

But first: I am going to assume you have adequate knowledge of functions and limits and such, and your main struggle in the topic of limits is to do with understanding the definition above. If you have no idea what limits are, then this is going to be a difficult read for you regardless of my explanation.

With that notice out of the way, let’s start the explanation.

Say I am convinced a limit exists for a certain function f(x) at x=a, with the limit called L. The function, by the way, is defined at every point on a certain interval (except perhaps at x=a). So I know there is a limit, and I know what it is. But how can I prove that I know the limit is L?

I can prove it as follows:

I challenge my knowledge with every possible value of “epsilon” that is greater than zero. Each value of “epsilon” represents a y-distance away from L (on either side of L). On the other hand, each value of “delta” represents an x-distance away from x=a, the x-coordinate that corresponds to L (on either side of x=a).

The challenge is to come up with a corresponding value of “delta” (in the case of each value of “epsilon”) that consistently keeps the portion of the graph within the x-distance of “delta” from x=a and within the y-distance of “epsilon” from L. The challenge fails if both aforementioned conditions are not met. The smaller the “epsilon”, the more challenging it is, since a small “epsilon” means a small distance away from L.

If, through logical thinking, you are able to deduce that every possible “epsilon” value has a corresponding “delta” value that works, then this means that no matter how close you want to get to the limit, there will always be a portion of the graph that will consistently stay within both the “delta” distance chosen and the “epsilon” distance it corresponds to. This shows that the graph is indeed approaching L as x approaches a, which means L is indeed the limit as x -> a.

And I think I’m done. If not, I’m still done anyway, because I’m tired.

[polymath257]

One way of understanding the epsilon-delta definition is in terms of tolerances.

Suppose you want to show that the limit of f(x) as x approaches a is L. Think of y=f(x) here.

We play a game: I give you a tolerance on y. That tolerance is called epsilon. This is how far from L I am willing to allow you to go.

You, in turn, have to give me a tolerance on x, which we call delta. This represents how far away from x=a you allow me to go.

Now, you win if *every* x value within your delta tolerance from 'a' produces y=f(x) value that is within my epsilon tolerance from L.

If, on the other hand, I can find an x that is *within* your tolerance, but where f(x) is *outside* of mine, then I win.

If the limit really is L, then you are guaranteed to win: you can always find a delta for any epsilon I pick.

I should say that understanding the epsilon-delta definition (and the corresponding epsilon-N definition for sequences) is a major step for those studying math. This one concept is what distinguishes a user of math from someone that actually understands it (well, up to a certain point). Almost everyone has trouble with it at first, so don't worry that all the 'for every, there exists, for every' stuff is tricky. Most people never have to go to that depth of quantifiers.

[robvalue]

An interesting feature is that limit of f(x) as x approaches t isn't always equal to f(t). In other words, the function can be converging on a value as you get arbitrarily close to a value of x, but it doesn't actually matter what happens exactly at the value, because you never get there. You can sometimes also get different results if you approach the limit from above or below.

For most functions people are familiar with however, the limit will be equal to the function at that point. These are continuous functions, which basically means there is no "jump" in their graph from one point to another.

Here is an example of a non-continuous function:

f(x) = x^2 for x not equal to 3
f(3) = 5

So this is a standard x^2 graph, except the value at 3 is 5 instead of the usual 9. There is a discontinuity there, a "jump". However, the limit of f(x) as x approaches 3 is 9, not 5.

[polymath257]

An interesting feature is that limit of f(x) as x approaches t isn't always equal to f(t). In other words, the function can be converging on a value as you get arbitrarily close to a value of x, but it doesn't actually matter what happens exactly at the value, because you never get there.  You can sometimes also get different results if you approach the limit from above or below.

For most functions people are familiar with however, the limit will be equal to the function at that point. These are continuous functions, which basically means there is no "jump" in their graph from one point to another.

Here is an example of a non-continuous function:

f(x) = x^2 for x not equal to 3
f(3) = 5

So this is a standard x^2 graph, except the value at 3 is 5 instead of the usual 9. There is a discontinuity there, a "jump". However, the limit of f(x) as x approaches 3 is 9, not 5.

Step functions are a good example of this. Tax rates are a good example: they suddenly jump at certain incomes (well, supposedly).

[robvalue]

Good point, yes

If we have f(x) = greatest integer less than or equal to x

Then f(1) = 1

Lim x->1 f(x) = 0 from below

Lim x->1 f(x) = 1 from above

I love maths. I asked for homework in it when I was 5.

[GrandizerII]

How about left-hand and right-hand limits:

Say, you have a function h(x) = sqrt(x). Since we're not dealing with imaginary numbers, the domain in this case would the set of all nonnegative numbers, so the least possible argument of the function would be 0.

What is the limit of h(x) as x approaches 0 from the right-hand side (the positive side)? Well, the function from the right to 0 is continuous and ends at the value of 0 at x = 0.

But what about the left-hand limit as x approaches 0 from the negative direction? Well, since we aren't dealing with negative x's in this case, then it seems the left-hand limit is undefined. Did I get this right? Or should I actually consider negative x's since they do lead to well-defined values (and, therefore, the limit would be 0 as well)? Or is the answer context-based?

[polymath257]

Good point, yes

If we have f(x) = greatest integer less than or equal to x

Then f(1) = 1

Lim x->1 f(x) = 0  from below

Lim x->1 f(x) = 1 from above

I love maths. I asked for homework in it when I was 5.

Technically, here, the limit doesn't exist. If the limit *does* exist, the only possibility is what you gave: a removable discontinuity.

And there are many ways a limit can fail to exist:

1. A jump discontinuity (as above)

2. A vertical asymptote (where the limit is infinite). F9x)=1/x does this as x->0.

3. Even the one-sided limits can fail to exist through oscillation. f(x)=sin(1/x) does this as x->0.

[robvalue]

How about left-hand and right-hand limits:

Say, you have a function h(x) = sqrt(x). Since we're not dealing with imaginary numbers, the domain in this case would the set of all nonnegative numbers, so the least possible argument of the function would be 0.

What is the limit of h(x) as x approaches 0 from the right-hand side (the positive side)? Well, the function from the right to 0 is continuous and ends at the value of 0 at x = 0.

But what about the left-hand limit as x approaches 0 from the negative direction? Well, since we aren't dealing with negative x's in this case, then it seems the left-hand limit is undefined. Did I get this right? Or should I actually consider negative x's since they do lead to well-defined values (and, therefore, the limit would be 0 as well)? Or is the answer context-based?

Yes, this is what I meant by "from above" and "from below", that's the same as right hand and left hand limits. Yours is probably better terminology!

If a function is defined only for 0 upwards, then it seems to me you can't take a left hand limit at 0, because there are no values to consider.

[polymath257]

How about left-hand and right-hand limits:

Say, you have a function h(x) = sqrt(x). Since we're not dealing with imaginary numbers, the domain in this case would the set of all nonnegative numbers, so the least possible argument of the function would be 0.

What is the limit of h(x) as x approaches 0 from the right-hand side (the positive side)? Well, the function from the right to 0 is continuous and ends at the value of 0 at x = 0.

But what about the left-hand limit as x approaches 0 from the negative direction? Well, since we aren't dealing with negative x's in this case, then it seems the left-hand limit is undefined. Did I get this right? Or should I actually consider negative x's since they do lead to well-defined values (and, therefore, the limit would be 0 as well)? Or is the answer context-based?

You got this right. The function is only defined (well, for real values) for x>=0.

In general, if we are looking at the endpoints of an interval, we test continuity using the appropriate one-sided limit.

The next interesting expansion of this is to consider functions of more than one variable. So, we can look at things like the limit of  f(x,y)=x^2+y^3 as (x,y)->(1,2).

The only modification is that in the 'x' tolerance, delta, we consider the points (x,y) that are closer to (1,2) than delta. So,

For every epsilon>0, there is a delta>0 such that for every (x,y) with dist( (x,y), (1,2) )<delta we get |f(x)-L|<epsilon.

We use the Pythagorean theorem to compute the distance: dist( (x,y), (1,2) )=sqrt{ (x-1)^2 + (y-2)^2 }

After this, the definition for functions of any number of variables is easy. *grin*

---

How about left-hand and right-hand limits:

Say, you have a function h(x) = sqrt(x). Since we're not dealing with imaginary numbers, the domain in this case would the set of all nonnegative numbers, so the least possible argument of the function would be 0.

What is the limit of h(x) as x approaches 0 from the right-hand side (the positive side)? Well, the function from the right to 0 is continuous and ends at the value of 0 at x = 0.

But what about the left-hand limit as x approaches 0 from the negative direction? Well, since we aren't dealing with negative x's in this case, then it seems the left-hand limit is undefined. Did I get this right? Or should I actually consider negative x's since they do lead to well-defined values (and, therefore, the limit would be 0 as well)? Or is the answer context-based?

Yes, this is what I meant by "from above" and "from below", that's the same as right hand and left hand limits. Yours is probably better terminology!

If a function is defined only for 0 upwards, then it seems to me you can't take a left hand limit at 0, because there are no values to consider.

Both terminologies are used. We also write lim x->0^+ for the limit from the right (above). Use a negative sign for the left (below).

[robvalue]

Good point, yes

If we have f(x) = greatest integer less than or equal to x

Then f(1) = 1

Lim x->1 f(x) = 0  from below

Lim x->1 f(x) = 1 from above

I love maths. I asked for homework in it when I was 5.

Technically, here, the limit doesn't exist. If the limit *does* exist, the only possibility is what you gave: a removable discontinuity.

And there are many ways a limit can fail to exist:

1. A jump discontinuity (as above)

2. A vertical asymptote (where the limit is infinite). F9x)=1/x does this as x->0.

3. Even the one-sided limits can fail to exist through oscillation. f(x)=sin(1/x) does this as x->0.

Yes, you are correct, the overall limit at X=1 does not exist because the two sides don't match.

Your example 3 is very interesting, I remember it from my lectures a long time ago. They would call it "not well behaved", informally.