Cannot decipher and solve limits help

1. Nov 5, 2004

semidevil

this is very elemetary, but i'm not following the logic at all. i'm trying to explain to myself after reading the book defn, but still, no luck.

Ok, so the defintion of limit is If Given e > 0, there exists a d > 0 such that if x belongs to A and 0 < |x- c| < d, then |f(x) - L| < e.

here, e = epsilon, and d = delta.

ok, so i'm explaining this to myself as, " if I let e > 0. I can find a delta > 0, that if x is in a, and x - c(with x not = c) is less then delta, then f(x) - L < e.

So when I do a problem, such as "the limit (from x to c) x = c, i'm having trouble how the book solves it.

the book does: let g(x) = x for all x in R. If e > 0, let delta = e. then if 0 < | x-c| <d, then |g(x) - c| = |x - c| < e. since e > 0, it proves it.

===================================

so I try to explain this to myself:

first, the book assigns g(x) = x.

then it goes through the definition. It lets e = d. so now, if 0 < |x - c| < d, then it means |x - c| < e.

so why does that prove it?

I dotn know why I can't get it...I think i'm missing something simple here.

2. Nov 6, 2004

StatusX

Just to be clear, this seems to be the definition of the following equation:

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

this is read as "the limit as x goes to c of f(x) equals L." heres an example to help you understand it. picture the function y=x^2. now its clear that as x approaches 0, y approaches 0 as well. in fact, in this example y is equal to 0 at x=0. but i could have said x^3/x, in which case the value at 0 is not strictly defined, although at all other points the function is the same. anyway, the epsilon delta definition is a way of formalizing this intuitive idea of a limit. think of a number e that is very very small, but still greater than 0. now picture all points on y=x^2 such that y is less than this value. now no matter how small this region is, it will always correspond to a small region around x=0. all you have to do is prove this region has some positive size d for any e you could pick. in this case, d would be sqrt(e), and this is greater than 0 as long as e is.

the last thing you said is actually what you need to prove. and x can be c, its just that in some limits the value of f(c) is not defined, such as sin(x)/x at x=0, but the limit still exists. let me know if this doesnt answer your question.

3. Nov 6, 2004

matt grime

If |x-c| is less than d and d is equal to e, then |x-c| is less than e too, and hence

|x-c|<d implies |g(x)-c| <e since g(x)=x, and d =e