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suneja

  • 2 years ago

lim x^2+x-11=9 x tends to 4 prove tis by formal definition of limit

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  1. Jeffrey_Calderon
    • 2 years ago
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    that's quite long to type. haha. you can do it dude! :)

  2. suneja
    • 2 years ago
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    sure i can ... wanted to jst chk

  3. heedcom
    • 2 years ago
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    just plug 4 into the equation

  4. nickhouraney
    • 2 years ago
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    @suneja using epsilon and delta right

  5. suneja
    • 2 years ago
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    @nickhouraney ya can u show dat here

  6. rerukumash
    • 2 years ago
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    wats lim anyone pls help

  7. suneja
    • 2 years ago
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    lim is limit

  8. nickhouraney
    • 2 years ago
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    ya i can show u

  9. rerukumash
    • 2 years ago
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    ok where is it used

  10. rerukumash
    • 2 years ago
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    pls reply

  11. nickhouraney
    • 2 years ago
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    |x-a| |x-4| |f(x)-L|<epsilon | (x^2+x-11) - 9| start from here

  12. satellite73
    • 2 years ago
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    start with \[|x^2-x-20|<\epsilon\] and see what you need for \(|x-4|\) to make it happen

  13. suneja
    • 2 years ago
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    mod (x^2-x-11)-9 < epsilon wenever 0<mod x-4 < delta tis is how v start then go by simplifying left inequality to find a suitable value for delta then verify that choice of delta i hav a prob in verification .. need help der

  14. nickhouraney
    • 2 years ago
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    we basically need to take this | (x^2+x-11) - 9| < epsilon and make it |x-4|<epsilon start by simplifying | (x^2+x-11) - 9| that will turn to | x^2+x-11 - 9| < epsilon

  15. suneja
    • 2 years ago
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    ya then u hav to prove that "choice " of delta wic u get by solving the 1st part like u said can u show me 2nd part of tis prob

  16. nickhouraney
    • 2 years ago
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    for the proof you would just go in reverse basically. if |x-4| < delta then |x-4| < (your choice for epsilon) then show how this will lead to |f(x)-L| < epsilon

  17. suneja
    • 2 years ago
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    @nickhouraney thanks

  18. nickhouraney
    • 2 years ago
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    well lets see your proof

  19. suneja
    • 2 years ago
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    i've done exactly like but was lil confused in 2nd half of it! neways thanks

  20. satellite73
    • 2 years ago
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    oh yes it is. okay you should start with \[|x^2+x-20|<\epsilon\] then factor to get \[|(x-4)(x+5)|<\epsilon\] you have control over \(|x-4|\) so you just need a bound for \(|x+5|\)

  21. satellite73
    • 2 years ago
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    trick is to say that since you are taking the limit as \(x\to 4\) you can assert that say \(3<x<5\) so the largest \(|x+5|\) can be is 10

  22. suneja
    • 2 years ago
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    wat do u mean by " a bound for \[\left| x+5 \right|\]

  23. satellite73
    • 2 years ago
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    ok lets go slow you get to say how large \(|x-4|\) can be, that is you get to pick your \(\delta\) so that \(|x-4|\) is smaller than \(\delta\)

  24. satellite73
    • 2 years ago
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    now you are looking at the product inside the square root, you have \(|(x-4)(x+5)|\) and you want to make this smaller than \(\epsilon\)

  25. satellite73
    • 2 years ago
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    but \(|(x-4)(x+5)|=|x-4||x+5|\) you can make the first term as small as you like, but you cannot make \(|x+5|\) small

  26. satellite73
    • 2 years ago
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    not square root, absolute value is what i meant

  27. satellite73
    • 2 years ago
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    so the question is, how big can \(|x+5|\) be? well it can be really really large, but don't forget you are making \(x\) close to 4

  28. satellite73
    • 2 years ago
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    so you can assert that since \(x\) is close to 4, it is certainly less that say 5, making \(|x+5|<10\)

  29. satellite73
    • 2 years ago
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    now you want \[|(x-4)(x+5)|<\epsilon\] if you make \(\delta=\frac{\epsilon}{10}\) and \(3<x<5\) then you know \[|(x-4)(x+5)|<10|x-4|<10\times \frac{\epsilon}{10}=\epsilon\]

  30. suneja
    • 2 years ago
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    got it ..thanks all

  31. suneja
    • 2 years ago
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    \[-\infty \rightarrow +\infty\] is not a bound right?

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