What is the Optimal Delta for a Given Epsilon and Limit?

[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 $$\frac{9}{503}$$

 

[tiny-tim]

x = $$\frac{-9}{497}$$

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

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.

 

[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?

 

[tiny-tim]

Hi ƒ(x)!

(just got up :zzz: …)

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

Because if it works for -δ₁ < x < δ₂, and δ₁ > δ > δ₂,

then it does not work for x = -δ, does it?

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

The definition says |x| < δ, so δ must be positive.

 

[ƒ(x)]

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

 

[tiny-tim]

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

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.

 

[ƒ(x)]

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

 

[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.