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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

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  1. zepdrix
    • one year ago
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    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
    • one year ago
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    \[\large\rm |3x-9|\lt0.01\]Do you see it? :) It's so close!!

  3. anonymous
    • one year ago
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    X < 3.003?

  4. anonymous
    • one year ago
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    sorry I suck at math lol

  5. zepdrix
    • one year ago
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    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
    • one year ago
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    So if I factor out a 3,\[\large\rm |3(\color{orangered}{x-3})|\lt0.01\]

  7. zepdrix
    • one year ago
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    And divide the 3 over to the other side,\[\large\rm \color{orangered}{|x-3|}\lt\frac{1}{300}\]

  8. zepdrix
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    cee peeeeeee +_+ where you at broski? brain esplode?

  11. anonymous
    • one year ago
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    thanks so much, i understand it a little better now

  12. zepdrix
    • one year ago
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    ya these are super weird :U calc gets a lot more fun once you get past this initial stuff

  13. anonymous
    • one year ago
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    i hope so lol

  14. anonymous
    • one year ago
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    can you help me with one more?

  15. zepdrix
    • one year ago
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    sure!

  16. anonymous
    • one year ago
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    thanks!given the epsilon-delta definition prove that lim(3x-5)=7 as x --> 4

  17. anonymous
    • one year ago
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    so would it become |x-4| < \[\epsilon \div 3\]

  18. anonymous
    • one year ago
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    so delta would be epsilon/3

  19. zepdrix
    • one year ago
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    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
    • one year ago
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    okay thanks!

  21. zepdrix
    • one year ago
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    \[\large\rm \lim_{x\to4}3x-5=7\] 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
    • one year ago
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    But ya you've got the right idea with finding the corresponding delta value :D

  23. zepdrix
    • one year ago
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    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
    • one year ago
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    Nothing to add, you've done a fine job @zepdrix :)

  25. zepdrix
    • one year ago
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    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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