- #1
KiwiKid
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Homework Statement
Prove that lim[x->5] 3x^2 - 1 = 74
Homework Equations
Epsilon-delta stuff.
The Attempt at a Solution
Ok, so first we have to find delta in terms of epsilon, and this is particularly sucky because it's a nonlinear function. Here's my attempt:
|3x^2 - 1 - 74| = |3||x - 5||x + 5| < ε → |x - 5||x + 5| < ε/3
Define C > |x + 5| → |x + 5||x - 5| < C|x - 5|
Define |x - 5| < ε/(3C) = δ
Now we have to find an actual value for C, so let's put a boundary of 1 on δ:
|x - 5| < 1 → |x + 5| < 11
C = 11
Now here's my attempt at the actual proof:
Let ε > 0, δ=min{1, ε/(3C)}=min{1, ε/33}
|3x^2 - 75| = |3||x + 5||x - 5| < 3 * 11 * ε/(3C) = 33 * ε/33 = ε
Therefore |x - 5| < δ implies |(3x^2 - 1) - 74| < ε
Ergo lim[x->5] 3x^2 - 1 = 74
Does that sound about right? I have the feeling that my proof might be missing a few bits. I also don't completely understand why I have to say δ=min{1, ε/33}. I suppose there's a specific reason δ=ε/33 won't work, but I don't see it.