1. ## Prove Limits Help

$\displaystyle \lim_{x\rightarrow 1} \frac{2+4x}{3} = 2$

$\displaystyle 0<\left | x-1 \right | < \delta$

$\displaystyle \left | \frac{2+4x}{3}-2 \right | <\epsilon$

simplified i get $\displaystyle \left | x-4 \right |< \frac{\epsilon }{4}$

i set $\displaystyle \frac{\epsilon }{4} = \delta$
then i get $\displaystyle 0< \left | x-1 \right | < \frac{\epsilon }{4}$

I dont have an answer key for that problem in my calculus book so i was wondering if i did it right?
Thankss

2. ## Re: Prove Limits Help

You want to prove $\displaystyle \left | \frac{2+4x}{3}-2 \right | <\epsilon$.

$\displaystyle \left | \frac{4x-4}{3} \right | <\epsilon$

$\displaystyle \left | \frac{4}{3} (x-1)\right | <\epsilon$

$\displaystyle | x-1 | <\frac{3}{4}\,\epsilon$. 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 epsilon-delta limit proof - it's in one of the "sticky" threads in this forum.

- Hollywood