# episilon delta proofs

G

#### Guest

##### Guest
How do you do an epislon delta proof for
$$\displaystyle \lim_{x\to4}\ (3x+4)=16$$?

Can someone help me and explain it to me? I don't understand it.

#### Gene

##### Senior Member
I'm not positive what an episilon delta proof is but I think you want
$$\displaystyle \lim_{dx\to0}\ (3(4+dx)+4)=16+e$$
12+3*dx+4=16+e
3*dx=e
as dx -> 0, e ->0
where dx is delta x and e is episilon.
If I'm wrong, (and no one else pops in) do you have an example from your book? It doesn't make much sense to me when you can just substitue x=4

G

#### Guest

##### Guest
Sorry. I don't have it. It's something that my teacher understands, and a few other mathematicans who are really smart to understand this advanced mathematical proof and ideology. lol. It has something to do with epsilon and delta though. =\.

#### soroban

##### Elite Member
Hello, atse1900!

How do you do an epislon delta proof for: $$\displaystyle \lim_{x\to4}\ (3x+4)=16$$?

Can someone help me and explain it to me? I don't understand it.
I won't go into the theory behind it, but here's the procedure.

The epsilon statement is: .$$\displaystyle |(3x\,+\,4)\,-\,16|\,<\,\epsilon$$ . [1]

. . and we must manipulate it into the form: .$$\displaystyle |x\,-\,4|\,<\,\delta$$ . [2]

We have: .$$\displaystyle |3x\,-\,12|\,<\,\epsilon$$

Factor: . . $$\displaystyle |3(x\,-\,4)|\,<\,\epsilon$$

. . . . . . . . . $$\displaystyle 3|x\,-\,4|\,<\,\epsilon$$

. . . . . . . . . . $$\displaystyle |x\,-\,4|\,<\,\frac{\epsilon}{3}$$

We have [2] if: $$\displaystyle \delta\,=\,\frac{\epsilon}{3}\qquad\leftarrow$$(This is the answer)