# Confused about limit of function

1. Oct 18, 2004

### semidevil

I'm not understanding the definition. For any epislon > 0, there exists a delta > 0, st if x belongs to A and 0 < |x - c| < delta, then |f(x) - L| < epsilon.

ok, so to solve a problem, it means that I assume that epsilon is greater then 0, and then, depending on the problem I need to pick a delta that is greater then 0, and it has to satisfy the 0 < |x - c| < delta....and then, it will satisfy |f(x) - L | < epsilon right?

I"m not quite understanding it. So here is the example from book

prove that the lim b from x to c = b.

So first, the book picks f(x) = b. So for any epislon > 0, they let delta be 1. this means if 0 < | x - c | < 1, then |f(x) - b| = |b - b| = 0, which is < epislon.

I dont follow this quite. how/why did they choose f(x) = b? I know it makes sense because b - b = 0, and then it is less then epsilon, so QED, but doesn't it mean you can do the same thing to all limits?

2. Oct 18, 2004

### NateTG

No. The limit of a constant function is trivial.

For example, for
$$f(x)=x$$
For $$\lim_{x\rightarrow 0} f(x)=0$$
$$\epsilon\leq\delta$$

if you tried $$\epsilon = 1$$ for some $$\delta < 1$$ it doesn't work.

3. Oct 19, 2004

### matt grime

"lim b from x to c = b."

try writing that more clearly.

f is the function whose limit you're trying to find at some point. it is unclear from what yo'uve written exactly what is going on.

click NateTG's maths symbols to see the latex for this kind of thing.

4. Oct 19, 2004

### MiGUi

Most people hates the epsilon delta's definition of a limit, but it is actually very simple to understand.

$$\forall \epsilon > 0 \quad \exists \delta > 0 : \left| x - a \right| < \delta \rightarrow \left| f(x) - L \right | < \epsilon$$

What is what in this expression?

Firstly, we have to say that 'a' is the point where we want to know the value of the function, but there the function is not defined, so we have to use the value of the function on x, which is very near to a but we have to demonstrate that the distance between x and a is already very very small.

We have to restrict the zone where we will find out the solution, so epsilon will be taken to restrict the vertical values of our function, there will be two horizontal lines: $$y = L \pm \epsilon$$. That is because we want to begin getting closer to the limit, and logically we wouldn't want to begin from infinity.

Then, we have that f(x) is a point in the vertical axe and L also is a point in that axe, so f(x) - L is a segment which gets lower as f(x) gets closer to L, the vertical coordinate of the point in which our function doesn't work. Thats it, f(x) begins getting closer to L from the higher point that epsilon lets.

The same can be applied to another segment |x - a| but this time is in the horizontal axe. As x gets closer to a, the segment gets lower and lower.

We can read then that as x gets closer to a, f(x) gets closer to L, and it is true always because for any epsilon exists at least one delta. And since delta and epsilon are very very tiny numbers, the value of the function in that point is L.

MiGUi

5. Oct 20, 2004

### spacetime

limit game

Let us see it like this.

You have a function $$f(x)$$, want to find its limit as $$x \rightarrow a$$. You make a claim that the limit is $$b$$.
For that, you need to show that you can make the difference $$|f(x)-b|$$ smaller than any positive number.

So , I give you a small number and ask you to make the difference smaller than that. By guessing and trying, you come up with a value of $$x$$
for which the function attains such a value as makes the difference smaller than the number I gave you. Not only that, you need to show that the difference is less for all the values of $$x$$ less than the value you got.

This is the idea. Now, you need to show that for any number I give you, you can do that.
So, in your post, the number corresponding to the one I gave you was $$\epsilon$$, the number you came up with was $$\delta$$, and for all 0 < |x - c| < delta means you have make the difference smaller than $$\epsilon$$ for all values less than the value you came up with.

Hope that helps.

spacetime
www.geocities.com/physics_all/index.html