Question

Limits are the devil

Answer 1

Can someone guide me in the right direction?

Answer 2

Epsilon Delta? Oooo these are nasty >.<
Lemme look at some notes.. maybe I can remember how to set this up.

Answer 3

The idea is,
For all positive epsilon, there exists a positive delta such that,

when x is within delta is 3,
our function (the thing we're taking the limit of) will be within epsilon of 15.

We set up the stuff, and try to turn our epsilon expression into the delta one.. and some info pops out.. blahh hold up .. gimme few minutes to research >.<

Answer 4

Yes your right they are nasty lol!!!!
thanks for helping me. I am self taught in this class.

Answer 5

\[\Large\rm \forall~ \epsilon \gt 0\quad \exists~\delta\gt0:\]In words, the above says:
`For all epsilon great than 0, there exists a delta greater than zero, such that`

If \(\Large\rm 0<|x-3|\lt\delta\), then \(\Large\rm |x^3-5x+3-15|\lt\epsilon\)

So when our x is some distance delta away from 3,
our function is some epsilon away from the limit.

Now what we try to do is get an x-3 out of our epsilon expression I think >.<
Hmmm

Answer 6

We're hoping that we can probably get an x-3 out of that expression.
So we should try long division, expecting it to work.