# Epsilon Delta Limites

1. Oct 30, 2009

### ƒ(x)

Given the limit of $$\frac{x^2+2x}{x^2-3x}$$ as x approaches 0 equals $$\frac{-2}{3}$$ and that ε = .01, find the greatest c such that every δ between zero and c is good. Give an exact answer.

0 < |x-0| < δ
0 < |$$\frac{x^2+2x}{|x^2-3x}$$ + $$\frac{2}{3|}$$| < ε

|$$\frac{x(x+2)}{|x(x-3)}$$ + $$\frac{2}{3|}$$| = .01
|$$\frac{x+2}{|x-3|}$$| = $$\frac{-197}{300}$$
300(x+2) = -197(x-3)
300x + 600 = -197x + 591
497x = -9
x = $$\frac{-9}{497}$$

But, my book got $$\frac{9}{503}$$

Last edited: Oct 30, 2009
2. Oct 30, 2009

### tiny-tim

Hi ƒ(x)!

The ε has to be valid on both sides of -2/3 (and positive) …

that gives you 9/497 and 9/503, and 9/503 is smaller.

3. Oct 30, 2009

### ƒ(x)

Ok, but why do I pick the smaller one? If the larger delta works then every delta smaller than that will also work, right?

Also, does it matter if its negative since there is an absolute value sign in the inequality?

4. Oct 31, 2009

### tiny-tim

Hi ƒ(x)!

(just got up :zzz: …)
Because if it works for -δ1 < x < δ2, and δ1 > δ > δ2,

then it does not work for x = -δ, does it?
The definition says |x| < δ, so δ must be positive.

5. Oct 31, 2009

### ƒ(x)

Ok, it's beginning to click. Could you give me a walk through of how you would do this problem?

6. Oct 31, 2009

### tiny-tim

I'd solve (x+2)/(x-3) = -2/3 ± 0.01 (isn't that what you did?), and take the smaller of those two |x|s.

7. Nov 1, 2009

### ƒ(x)

Ok. Can you explain again why you take the smaller of the two values?

8. Nov 1, 2009

### tiny-tim

You've found f(-δ1) = f(δ2) = ± ε.

Now you need a δ such that if |x| < δ, then |f(x)| < ε.

If δ1 > δ2, and if we choose δ = δ1, then the statement "if |x| < δ1, then |f(x)| < ε" isn't true, because although f(-δ1) = ε, f(δ1) > ε.