# Understanding formal definition of limits

1. Nov 25, 2006

So far I could not understand the formal definition of limits. I know that limit of a function exists at a point say 'a' if left and right hand limit exist at that point. Then what is the need for the formal definition of limits?
Secondly, I am very confused by the usage of the notation of delta and epsilon and their usage in that definition.

2. Nov 25, 2006

### mathwonk

that is defining something in terms of itself, considered circular. defining intelligence as the quality possessed by someone who is intelligent.

to define a limit you may not use the word limit in the definition. the point is to describe it to someone who does not understand the meaning of the word limit.

you have acomplished something, namely you have defiend a 2 sided limit for someone who already knows what a one sided limit is. now define a one sided limit and you will be done.

3. Nov 25, 2006

### mathwonk

oops actually your definition is wrong,a s the 2 one sided limits must exist AND be equal.

4. Nov 25, 2006

### quasar987

5. Nov 25, 2006

### Hubert

I didn't want to start a new thread, but I have a question of my own:

I think essentially what the limit definition says is that
$$\lim_{x\rightarrow a} f(x) = L$$

means for each $\epsilon>0$ there is a $\delta>0$ such that f(x) is in the inverval $(L-\epsilon, L+\epsilon)$ whenever x is in the the interval $(a-\delta,a+\delta)$ (x does not equal a).

Now, what I am wondering is: do the upper and lower bounds of the interval have to be the same distance from L and a?

Or could there be two epsilons and two deltas, making the limit definition:

$$\lim_{x\rightarrow a} f(x) = L$$

means for each $\epsilon_1>0$ and $\epsilon_2>0$, there are $\delta_1>0$ and $\delta_2>0$ such that f(x) is in the inverval $(L-\epsilon_1, L+\epsilon_2)$ whenever x is in the the interval $(a-\delta_1,a+\delta_2)$ (x does not equal a).

Could this definition also work?

Last edited: Nov 25, 2006
6. Nov 25, 2006

The second 'definition' wouldn't make sense, unless I'm missing something, since the first states that for each $\epsilon>0$ ... So there must be more than one $\epsilon>0$.