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I am very good with limits but I am having a hard time solving hard exercises with limit definition.

Can anyone provide resource where I can (in my limited time) understand it well?

MATH000
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  • Do you mean limits of sequences or limits of functions? For sequences, consider the problem book Sequences, Combinations, Limits by Gelfand et al. – David Feb 17 '16 at 16:18

2 Answers2

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If you think about it in terms of neighborhoods, it becomes a lot simpler. A neighborhood of a point is a set of points "right around it," basically a bubble of numbers containing the point. For example, a neighborhood of the point 0 is any interval of the form (-a,a). To say that a function approaches a limit as the input goes to a certain point means that the function value gets close to some eventual value (the limit) as the input gets close to the point. This is made rigorous by saying that for every bubble around the limit value (neighborhood of the limit) we can put the function value in that bubble if we only squeeze the input value within some bubble around (neighborhood of) the point we are taking the limit at. The epsilon just corresponds to a neighborhood of the limit value, while the delta corresponds to a neighborhood of the point at which we are taking the limit. In doing an epsilon-delta proof, you are given a bubble around the limit value (epsilon) and have to find a bubble around the point at which you are taking the limit (delta) in which the function value lies close enough to the limit value to fall inside the epsilon-bubble. It then becomes a bunch of algebra. Alternatively, try http://tutorial.math.lamar.edu/Classes/CalcI/DefnOfLimit.aspx.

Davey
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If you mean the definition:

Given $\epsilon >0$ there exists $N$ such that for all $n>N$ we have $|x_n -a| <\epsilon$.

this means that the values of your sequence can be made to come as close as you wish to the limit value $a$ by looking far enough along in your sequence.

For example consider the sequence $$x_n = \frac{1}{n}$$

Suppose I want to the sequence to be at most $1$ unit away from the limit value $0$. Here we are effectively setting $\epsilon = 1$ in the above definition. To do this it will be enough to choose the vale of $N$ to be $1$, since for all $n>N=1$ $$|x_n -0| = \bigg |\frac{1}{n}\bigg| = \frac{1}{n}<1$$ We are now over the moon since we have found an $N$ that works for $\epsilon = 1$. But our excitement is short-lived.

In order to actually prove what we want, we need to do what we just did above, however this time we need to do it for all $\epsilon>0$.

If we can succeed in doing this, then what we have shown is I can make the terms of my sequence come however close I want to the actual value of the limit!

fosho
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