Thread: Epsilon - Delta Definition of Limit.

We want . This is the case if . We have an upper bound for the numerator: . To bound from above, try bounding from below when .

Hey powercroat783.

For these kind of problems, you want to get an epsilon in terms of a delta. So you have a bound for x in terms of delta, and now you want to get an epsilon in terms of that delta so that when you use the epsilon relation above (i.e. |blah| < epsilon) the x's are in terms of epsilons (or at least some algebraic expression is) and then you show that the stuff in the expression meets the criteria with the given epsilon.

So the real thing for these problems becomes picking a epsilon that is a function of delta and you know that |x+2| is in-between 0 and delta. So you have 3x+6 = 3(x+2) > 3x+2 so you want to consider your epsilon in terms of this function.

The triangle inequality is useful in that |A +- B| <= |A| + |B| < epsilon.

Usually the way that this kind of thing is done is that if you have the A and B terms above, you want to show that each term is less than epsilon/2 and then you use this triangle inequality to bring things together. So try and focus on getting the larger term less than epsilon/2 and show that the other ones is less than that and you're done.

Originally Posted by chiro

For these kind of problems, you want to get an epsilon in terms of a delta. So you have a bound for x in terms of delta, and now you want to get an epsilon in terms of that delta so that when you use the epsilon relation above (i.e. |blah| < epsilon) the x's are in terms of epsilons (or at least some algebraic expression is) and then you show that the stuff in the expression meets the criteria with the given epsilon.

I am having trouble understanding this, in particular phrases like "the x's are in terms of epsilons" and "the stuff in the expression meets the criteria with the given epsilon." Of course, in the end you want a delta in terms of an epsilon, not the other way around.

When I did these kind of things a long time ago, we got the expression regarding |f(x) - f(l)| < epsilon by showing that the expression inside the norm had those properties and thus got a condition of the terms with respect to epsilon.

I understand that all this is doing is saying that mult-dimensionally, you shrink the range of the domain around a point and the norm of the mapping between the two values shrinks as well intuitively, but your delta's can be written in epsilons as well as the epsilons written in delta.

I'm just explaining how I used to do this: you get a way of showing the norm of the difference of the final map has the properties by relating them to an expression involving epsilon and showing that these quantities maintain the inequality of the whole norm (i.e. your |f(x) - l| term).