(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Prove that lim[x->5] 3x^2 - 1 = 74

2. Relevant equations

Epsilon-delta stuff.

3. 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.

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# Homework Help: One more try at epsilon-delta limits

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