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Math Help - Answer Verication Needed For Infinity Limits Question

  1. #1
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    Answer Verication Needed For Infinity Limits Question

    I've solved these two questions involving limits approaching infinity but I'm not sure if they're right. Could someone verify them if they have time?

    -thanks

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  2. #2
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    Quote Originally Posted by SportfreundeKeaneKent View Post
    I've solved these two questions involving limits approaching infinity but I'm not sure if they're right. Could someone verify them if they have time?

    -thanks

    They are both correct. Though that's not how I would write them out, but
    if that's what you have been taught its what you have been taught.

    RonL
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by SportfreundeKeaneKent View Post
    I've solved these two questions involving limits approaching infinity but I'm not sure if they're right. Could someone verify them if they have time?

    -thanks

    i don't think that's how he's been taught Captain. if so, they should fire his professor.

    here's what's wrong with the way you wrote them out.

    1) you have to cancel the x's you factor out before evaluating the limit
    2) once you evaluate the limit, you drop the limit notation.

    in question 4 for instance, you did the following:

    lim{x-->infinity} (x^3 - 3x^2 - x + 4)/(5x^3 + 1)
    = lim{x-->infinity} [x^3(1 - 3/x - 1/x^2 + 4/x^3)]/[x^3(5 + 1/x^3)]
    = lim{x-->infinity} [x^3(1 - 0 - 0 + 0)]/[x^3(5 + 0)]
    = lim{x-->infinity} x^3/(5x^3)
    = 1/5


    what you should have done is this:
    lim{x-->infinity} (x^3 - 3x^2 - x + 4)/(5x^3 + 1)
    = lim{x-->infinity} [x^3(1 - 3/x - 1/x^2 + 4/x^3)]/[x^3(5 + 1/x^3)]
    = lim{x-->infinity} (1 - 3/x - 1/x^2 + 4/x^3)/(5 + 1/x^3)
    = (1 - 0 - 0 + 0)/(5 + 0) ........now i finally took the limit, so i drop the notation
    = 1/5

    did you notice the difference? it may not seem like a big deal, but in a way it is. it's like you were solving x + 5 = 7 in the following way

    x + 5 = 7
    => x = 7 + 5
    => x = 2

    sure you got the right answer, but its not the answer you should get based on what you wrote
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  4. #4
    Junior Member frenzy's Avatar
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    Shouldn't the answer to #5 be -infinity
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  5. #5
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    Ah you're right, number 5 should be negative infinity because of the two in front.

    And why do you have to cancel out the x? My high school math teacher doesn't show the work for most of these questions, I'm just basing it on stuff I found around the interent.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by SportfreundeKeaneKent View Post
    Ah you're right, number 5 should be negative infinity because of the two in front.

    And why do you have to cancel out the x? My high school math teacher doesn't show the work for most of these questions, I'm just basing it on stuff I found around the interent.
    all the x's in the function represent the same number, the x's you factor out are indistinguishable from the x's inside the brackets in other words. so once you take the limit as x goes to infinity, you have to let ALL the x's go to infinity. you can't let some x's go to infinity and others (the ones you factored out) don't. but if you let those x's go to infinity before you cancel them out algebraically, you end up with infinity/infinity --- which was the problem in the first place. to avoid that, you get rid of the x's that would cause the function to go to infinity/infinity first, then take the limit.

    got it?
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