# Homework Help: Proving the limit #2

1. Oct 8, 2009

### zeion

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. The attempt at a solution

0 < |(x + 3 / 2x - 1) - 4| < e
0 < | x+3-4(2x-1) / 2x-1 | < e
0 < | -7x+7 / 2x-1 | < e
0 < |-7(x-1) / 2x-1| < e

Not sure what to do now.

0 < ||-7||x-1|| / |2x-1| < e

How do I relate the x-1 with d?

2. Oct 9, 2009

### tiny-tim

Hi zeion!
oooh, that's so complicated!

just write the original numerator as a multiple of 2x-1, plus a remainder …

then it'll be obvious!

3. Oct 9, 2009

### zeion

You mean to write x+3 as a multiple of 2x-1?

4. Oct 9, 2009

### HallsofIvy

He means divide x+3 by 2x- 1: 2x+ 1 divides into x+ 3 "1/2" times with a remainder of 7/2:
$$\frac{x+3}{2x-1}= \frac{1}{2}+ \frac{\frac{7}{2}}{2x-1}$$
though I honestly don't see how that simplifies a lot.

$$0< 7\frac{|x-1|}{|2x-1|}< \epsilon$$