Confused about the definition of a limit

[torquerotates]

So the definition states.( For any epsilon>0, there exists a delta>0 such that 0<|x-a|<delta => |f(x)-L|<epsilon. ) is equivalent to lim f(x)=L for x->a.

Well say f<=C some constant.
(reads less than or equal to)

then L<=C But that means that the epslion neighborhood can't extend past C. f is always going to be in (L-epsilon, L+epsilon). If L+epsilon>C, then there exists x* such that f(x*)>C. This puts a restriction on the values of epsilon. But the definition states that for ANY epsilon>0. epsilon should be as great as I want it. But it isn't.

 

[disregardthat]

It puts no restriction on the epsilons. As you can see, if there is a delta > 0 such that |f(x)-f(y)|<epsilon when |x-y|<delta, then this delta works for all epsilons bigger than the one we have used! All we need is that the function is close enough to f(x) around x, so if we "give it more slack" to vary, it won't do any harm.

In your example it is no reason for that the epsilon neighborhood can't exceed the range of the function. If so you would e.g. immediately get in trouble with constant functions. As explained above larger epsilons only give weaker conditions.

 

[torquerotates]

Oh I see. But then what about smaller epsilon? Say |xsin(1/x)|<=|x|<epsilon? Can I say that even though epsilon is bounded below by |x|, since |x|<delta=epsilon, and delta>0, |x| can take on all values less than the current epsilon so we can find a epsilon smaller than the one we have but greater than |x|?

 

[HallsofIvy]

"even though epsilon is bounded below by |x|" is incorrect. |xsin(1/x)|<= |x|< epsilon puts a condition on x, not on epsilon. epsilon is "given" initially. You control values of x, not epsilon.