# Definition of a Limit

1. Nov 14, 2013

### alexjean

The problem statement, all variables and given/known data

I know what a limit is and I understand the idea behind it, but I am misunderstanding something in the formal definition of a limit.

DEFINITION
Let f(x) be defined on an open interval about x0, except possibly at x itself. We say that the limit of f(x) as x approaches x0 is the number L and write

limx → x0 f(x) = L

if, for every number ε > 0 there exists a corresponding number δ > 0 such that for all x,
0 < abs(x – x0) < δ
=>
abs(f(x) – L) < ε

The attempt at a solution

For example, take f(x) = x2

I could, incorrectly, assert that limx → 2 f(x2) = 1

Now, δ > abs(x - 2) > 0
and ε > abs(f(x) - 1)

If:
1) this holds true for all values of x (except x = x0 = 2), and,
2) for every possible value of epsilon some value of x which satisfies the above statement for both ε and δ exists,
Then:
1 is the limit of x2 as x approaches 2.

As an example, I pick x = 3. So,
δ must be > 3
and
ε must be > 8

I can continue and choose arbitrary values of x, none of which seem to be a problem.

Likewise, for any value of epsilon I can think of, I can find an value of x which satisfies the inequality ε > abs(f(x) - 1) which also satisfies δ > abs(x - 2) > 0.
ex: when ε > .01 I could have x = .999. This x also satisfies δ > abs(x - 2) > 0
(as far I can tell, it's impossible not to satisfy gamma, as long as x isn't equal to x0)

By that reasoning, the claim I made above is true and limx → 2 f(x2) = 1

By the reasoning I'm using with the formal definition, I can state that any value of L is the limit for any function at an arbitrary x0 (given that the function has values infinity close to L at some point in time). This is clearly incorrect. Would anyone please point out the error in my application of the definition of a limit?

Thank you.

2. Nov 14, 2013

### Staff: Mentor

Your error is that someone else chooses $\epsilon$, and from that you have to find a δ > 0 so that, when |x - 2| < δ, |f(x) - L| < $\epsilon$.

Using your example, where you postulate that $\lim_{x \to 2}x^2 = 1$, I will say that $\epsilon = .01$. You now have to find a pos. number δ such that for each x $\in$ (2 - δ, 2 + δ), then x2 $\in$ (.99, 1.01).

Sketch a graph of the function y = x2, and you'll see that this can't happen.