Question

HELP ME!!!!!!!!! Calculus! Do a delta-epsilon proof for:

Answer 1

\[\lim_{x \rightarrow 4}\frac{ x^2-16 }{ x-4} =8\]

Answer 2

Do you even need to do a delta epsilon proof? The top is (x-4)(x+4), so you're simply asked for the lim x->4 of x+4. That's pretty clearly 8...

Answer 3

@elementalmagicks Wow you're confused, lol.

Answer 4

i know but my teacher wants us to do it using a delta-epsilon proof :/

Answer 5

\[
|x-4|<\delta\implies \left|\frac{x^2-16}{x-4}-8\right| <\epsilon
\]

Answer 6

I'm confused apparently. wio hath spoken.

Answer 7

@voltron21 What is stopping you from finishing the proof?

Answer 8

@wio im as confused with this whole concept as @elementalmagicks

Answer 9

@elementalmagicks Yeah you are. There is a difference between a "proof" and just "evaluating"

Answer 10

\[
|x-4|<\delta\implies \left|\frac{x^2-16}{x-4}-8\right| <\epsilon
\]Becomes: \[
|x-4|<\delta\implies \left|\frac{(x-4)(x+4)}{x-4}-8\right| <\epsilon
\]Since it is a limit, we assume \(x\neq 4\) so we have:\[

|x-4|<\delta\implies \left|(x+4)-8\right| <\epsilon
\]

Answer 11

@voltron21 Is it making sense yet?

Answer 12

@wio kind of, so u set the restrictions and now you're simplifying

Answer 13

I suppose I feel silly giving an "epsilon delta" proof when delta can just be chosen to be epsilon.

Answer 14

Typically you have to choose more creatively, but this apparently was just asking you for the basic mechanics.

Answer 15

The idea of epsilon delta proofs is that epsilon is fixed and then you pick a delta such that the implications that wio was been writing as satisfied.

Answer 16

are*

Answer 17

@voltron21 Yes, and with these sorts of proofs you generally want to get to something of the form: \[
|x-a| <\delta \implies g(x)|x-a|<\epsilon
\] In this case \(g(x)=1\) so it is relatively easy \(\delta \sim \epsilon\).
If \(g(x)=c\) (a constant) we'd pick \(\delta \sim \epsilon/ c\)

Answer 18

@wio so is that it? is that the proof?

Answer 19

I'm sort of explaining the methodology for the proof.

Answer 20

@wio so ur not done?lol

Answer 21

The proof was done a while ago.

Answer 22

Once you get to:\[
|x-4|<\delta \implies |x-4|<\epsilon
\]You just say \(\delta \sim \epsilon\) and you're done.

Answer 23

@wio i thought i was supposed to get to the whole thing =epsilon

Answer 24

You have to basically let \(\delta\) be a function of \(\epsilon\), which we did.

Answer 25

@wio what does that little squiggle mean... sorry i dont understand math talk

Answer 26

@wio and where did the 8 go?

Answer 27

Don't worry about it, it doesn't have any special meaning.

Answer 28

\(=\) here is fine as well

Answer 29

@wio oh so it just means =

Answer 30

\(\sim\) means in the ballpark of, technically delta doesn't have to equal epsilon since it can be less.

Answer 31

@wio so it means about ok that makes sense but i still don't understand what happened to the 8

Answer 32

\[
x+4-8=x-4
\]

Answer 33

@wio oh lol got ya i just didnt see that...thanks!