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Homework Help: Limit of function

  1. Feb 10, 2010 #1
    1. The problem statement, all variables and given/known data
    1) limit of [ sqrt (1+2x) - sqrt (1-3x) ] / x as x tends to 0
    2) limit of (3x^4 -8x^3 + d) / (x^3 - x^2 - x + 1) as x tends to 1
    3) limit of [ sqrt (x^2 + 1) - sqrt (x^2 - 1) ] as x tends to infinity


    2. Relevant equations



    3. The attempt at a solution
    1) a little confusion here... is the limit 0 or 5/2?
    2) since the function is continuous at x0=1, then sub. 1 into all x and the limit does not exist since the denominator = 0 (is it correct?)
    3) divide each term inside the sqrt by x^2 and get sqrt 1 - sqrt 1 = 0
     
  2. jcsd
  3. Feb 10, 2010 #2

    Mark44

    Staff: Mentor

    Multiply by 1 in the form of [sqrt(1 + 2x) + sqrt(1 - 3x)] over itself.
    It depends on what d is. Or is that a typo?
    No, that won't work, because what you have now is x(sqrt(1 + 1/x^2) - sqrt(1 - 1/x^2)). This is indeterminate as x --> infinity. The thing to do is to multiply by 1 in the form of the conjugate over itself.
     
  4. Feb 10, 2010 #3
    1) yeah that's what I did and get 5/2 but I thought the answer would be 0
    2) yes it is a typo, it should be 5 instead of d. but the limit does not exist right? since the denominator = 0
    3) oh! mupltiply by (sqrt (x^2 + 1) + sqrt(x^2-) ) expand and get x^2 + 1 - (x^2 -1) = 0?
     
  5. Feb 10, 2010 #4

    Mark44

    Staff: Mentor

    For 1, if you show your work, I'll check it.
    For 2, the limit is -oo.
    For 3, x^2 + 1 - (x^2 -1) != 0. Check your algebra!
     
  6. Feb 10, 2010 #5
    1) multiply the equation by (sqrt (1+2x) + sqrt (1-3x)) / (sqrt (1+2x) + sqrt (1-3x))
    and get 5x / x (sqrt (1+2x) + sqrt (1-3x))
    and get 5 / (sqrt (1+2x) + sqrt (1-3x))
    sub. 0 into x and get 5/2

    2) how did the limit turn to be negative infinity???

    3) oops I'm sorry. I just realized it when I did it on the paper AFTER I posted that.
    if I multiply the whole equation by (sqrt (x^2 + 1) + sqrt (x^2-1)) / (sqrt (x^2 + 1) + sqrt (x^2-1) ), get
    2 / (sqrt (x^2 + 1) + sqrt (x^2-1) )
    then divide each term in the sqrt by x^2?
    and get 2/infinity = 0?

    does 0 / infinity = 0?
    and what is 0 x infinity?
     
    Last edited: Feb 10, 2010
  7. Feb 10, 2010 #6

    Mark44

    Staff: Mentor

    Right, the limit is 5/2. On a technical note, you're not multiplying an equation by that stuff; you're multiplying an expression by that stuff. An equation always has an = sign lurking about.
    Actually, my answer was incorrect. The limit doesn't exist at all. The left-side limit is oo and the right-side limit is -oo. To see what's happening around x = 1, notice that you can factor (x - 1) out of both the numerator and denominator.
    Prob 3 is pretty clear-cut. After multiplying the expression by the conjugate over itself, you get 2/(sqrt(x^2 + 1) + sqrt(x^2 - 1)). As x --> oo, the denominator gets large without bound, but the numerator is stuck at 2, so the whole fraction approaches 0.
     
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