## anonymous one year ago Given the lim(3x-7)=2 as x-->3. Using the epsilon-delta definition for the given epsilon=.01 find the corresponding delta > 0 such that |(3x-7) - 2| < .01 whenever |x-3| < delta

1. zepdrix

So when $$\large\rm \epsilon=0.01$$, there exists a $$\large\rm \delta\gt0$$ such that$\large\rm 0\lt|x-3|\lt\delta\qquad\implies\qquad |(3x-7)-2|\lt0.01$ So start with your epsilon equation, trying to find an x-3 in there.

2. zepdrix

$\large\rm |3x-9|\lt0.01$Do you see it? :) It's so close!!

3. anonymous

X < 3.003?

4. anonymous

sorry I suck at math lol

5. zepdrix

Well you're not trying to solve for x in this inequality! :O careful!$\large\rm \color{orangered}{|x-3|}\lt\delta$You're trying to make this orange part show up in your epsilon equation.

6. zepdrix

So if I factor out a 3,$\large\rm |3(\color{orangered}{x-3})|\lt0.01$

7. zepdrix

And divide the 3 over to the other side,$\large\rm \color{orangered}{|x-3|}\lt\frac{1}{300}$

8. zepdrix

If you compare these two inequalities:$\large\rm \color{orangered}{|x-3|}\lt\delta\qquad\qquad\qquad \color{orangered}{|x-3|}\lt\frac{1}{300}$It looks like we've found our delta, do you see it? :o

9. zepdrix

I rewrote 0.01 as 1/100 and then divided by 3, or multiplied by 1/3, just in case that middle step was confusing.

10. zepdrix

cee peeeeeee +_+ where you at broski? brain esplode?

11. anonymous

thanks so much, i understand it a little better now

12. zepdrix

ya these are super weird :U calc gets a lot more fun once you get past this initial stuff

13. anonymous

i hope so lol

14. anonymous

can you help me with one more?

15. zepdrix

sure!

16. anonymous

thanks!given the epsilon-delta definition prove that lim(3x-5)=7 as x --> 4

17. anonymous

so would it become |x-4| < $\epsilon \div 3$

18. anonymous

so delta would be epsilon/3

19. zepdrix

Mmm ya that sounds right! :) If they say "prove" then maybe we need to write it up formally, with some words and stuff...

20. anonymous

okay thanks!

21. zepdrix

$\large\rm \lim_{x\to4}3x-5=7$ Let $$\large\rm \epsilon\gt0$$. There exists a $$\large\rm \delta\gt0$$ such that $\large\rm \text{if}\quad 0\lt|x-4|\lt\delta\qquad\text{then}\qquad |f(x)-7|\lt\epsilon$ $\large\rm |f(x)-7|\lt\epsilon \quad\text{if and only if}\quad |3x-5-7|\lt\epsilon$$\large\rm \text{if and only if}\quad|3x-12|\lt\epsilon$$\large\rm \text{if and only if}\quad|x-4|\lt\frac{\epsilon}{3}$ Choose $$\large\rm \delta=\dfrac{\epsilon}{3}$$. Thus, $$\large\rm 0\lt|x-4|\lt\dfrac{\epsilon}{3}$$ and it follows that $$\large\rm |f(x)-7|\lt\epsilon$$. Maybe something like that? :p Bah I can't remember how to write these properly.

22. zepdrix

But ya you've got the right idea with finding the corresponding delta value :D

23. zepdrix

The first type of problem is a little easier to deal with, you're just looking for a specific number. In the second problem you're generalizing for every epsilon, so you gotta be careful :O

24. anonymous

Nothing to add, you've done a fine job @zepdrix :)

25. zepdrix

What do you think cee pee? :o Too much with the words? It's just each step following the previous step. Relating delta to the epsilon/3 is really the important part though, which it seems like you've figured out.

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