Proving Limit #2: 0 < ||-7||x-1|| / |2x-1| < e

[zeion]

Homework Statement

<br /> \lim_{x\rightarrow 1} \frac{x + 3}{2x - 1} = 4<br />

Homework Equations

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?

 

[tiny-tim]

Hi zeion!

Homework Statement

<br /> \lim_{x\rightarrow 1} \frac{x + 3}{2x - 1} = 4<br />

Homework Equations

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?

oooh, that's so complicated!

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

then it'll be obvious! ​

 

[zeion]

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

 

[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.

Since you have already done the work, go ahead with
0&lt; 7\frac{|x-1|}{|2x-1|}&lt; \epsilon
You just need a bound on 7/|2x-1|.

Start by requiring that |x-1|< 1/4 so that -1/4< x- 1< 1/4 and 3/4< x< 5/4. Then 3/2< 2x< 5/2 so 1/2< 2x-1< 3/2. That tells you that 1/2< |2x-1|< 3/2.
(I started with |x-1|< 1 but had to lower to 1/3 to keep those numbers larger than 0!)

Now you know that 2/3< 1/|2x-1|< 2 and so that 7|x-1|/|2x-1|< 14|x-1|.