So the problem is to prove lim┬(x→3)⁡〖(x+1)/(x-1)〗=2 (sorry if it looks weird, I copy and pasted it from Word)
I'm pretty sure I'm supposed to use the epsilon delta definition of limits to prove it. Our book doesn't tell us how to do that, but our teacher expects us to know it, so I looked it up online.
I set it up like this: For all ε>0, there is a δ>0 such that if |x-3|<δ, then |(x+1)/(x-1)-2|<ε.
Then I manipulated the |(x+1)/(x-1)-2|<ε to eventually get |(x-3)/(x-1)|<ε, which seems like it's on the right track to defining δ since |x-3|<δ. But if I use it like that then I get |x-3|<|x-1|ε, which would mean I would have to make δ=|x-1|ε and I'm pretty sure I'm not supposed to have x in the definition for δ.
I just need some help figuring out what I'm supposed to do to get rid of that |x-1| so I can define δ. I think once I define δ I can do the rest of the problem just fine, I'm just having trouble getting there.

sorry guys it came out looking really weird.
lim(x->3) (x+1)/(x-1) =2
ε is supposed to be epsilon
δ is supposed to be delta

Oh my goodness. AZA is the I with the hat.

Perhaps you mean this? Show that


\displaystyle \lim_{x \to 3}( \frac {x+1}{x-1}) = 2


you need to find \delta \gt 0 such that if

|x-3|<\delta

then

|\frac {x+1}{x-1}-2| \lt \epsilon


for any positive \epsilon .

Is that it?

Yes that's it, thanks for making it look better

I would do this a bit differently. Starting with Galactus's observation that for this problem |f(x)-L| = |(x-3)/(x-1)|, set:


| \frac {x-3} {x-1}| \lt \epsilon


We need to replace x with x \pm \delta and solve for positive values of \delta. This will show us that no matter what value of \epsilon is chosen, we can find a \delta that satisfies the required condition.


|\frac{(x\pm\delta)-3} {(x \pm \delta)-1}| = \epsilon


Set x = 3:

|\frac {\delta}{2\pm\delta} | \lt \epsilon


Note that both numerator and denominator are positive values (as long as \delta isn't too big), so we can drop the absolute value signs and solve for \delta :


\delta \lt \frac {2\epsilon} {1 \pm \epsilon}


Now you can pick any positive value for \epsilon , no matter how small, and show that if x is within \pm \delta of 3, the value of f(x) is within \epsilon of 2. So for example if \epsilon is picked to be 0.1, then \delta\ \lt \ 0.222 satisfies the condition. Hence


\displaystyle \lim_{x \to 3}( \frac {x+1}{x-1}) = 2

I tried giving you a rep, ebaines, but of course I have not spread it around enough :rolleyes:

Anyway, what I was trying to say is that your method is best. To the point and a good explanation of what is going on.

But, here is a graphical representation that may help you envision it.

Here is a epsilon delta applet that is very nice. Check it out.

Sample Configurable JCM Applet: EpsilonDelta (http://math.hws.edu/javamath/config_applets/EpsilonDelta.html)