
Prove Limits Help
$\displaystyle \lim_{x\rightarrow 1} \frac{2+4x}{3} = 2 $
$\displaystyle 0<\left  x1 \right  < \delta $
$\displaystyle \left  \frac{2+4x}{3}2 \right  <\epsilon $
simplified i get $\displaystyle \left  x4 \right < \frac{\epsilon }{4} $
i set $\displaystyle \frac{\epsilon }{4} = \delta $
then i get $\displaystyle 0< \left  x1 \right  < \frac{\epsilon }{4} $
I dont have an answer key for that problem in my calculus book so i was wondering if i did it right?
Thankss

Re: Prove Limits Help
You want to prove $\displaystyle \left  \frac{2+4x}{3}2 \right  <\epsilon $.
$\displaystyle \left  \frac{4x4}{3} \right  <\epsilon $
$\displaystyle \left  \frac{4}{3} (x1)\right  <\epsilon $
$\displaystyle  x1  <\frac{3}{4}\,\epsilon $. So it looks like it would work to set $\displaystyle \delta=\frac{3}{4}\,\epsilon $.
Now you need to write out the proof....
There's a really good paper describing how to construct an epsilondelta limit proof  it's in one of the "sticky" threads in this forum.
 Hollywood