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Limit question

  1. Oct 23, 2007 #1
    The problem is

    The limit as x approaches pos infinity ln(square root of x + 5) divided by ln(x)

    In the numerator only x is under the square root. I'm having trouble getting to this answer. If someone can take a look I would really appreciate it.
  2. jcsd
  3. Oct 23, 2007 #2


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    remember ln(sqrt(x)) = 1/2 ln(x).
  4. Oct 23, 2007 #3
    well you can multiply by sqr. rt of x -5/ sqr rt of x -5

    that leaves you with ln x-25/ x(sqr rt of x + 5)
  5. Oct 24, 2007 #4

    Gib Z

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    mathwonk's hint intentionally disregarded the 5 within the argument of the log. Intuitively, as x grows large the 5 within the log becomes insignificant and can be ignored. More rigorously, the natural log of (sqrtx + 5) is asymptotic to log (sqrtx), which means that the difference of the two for a given value of x goes to zero as x goes to infinity, basically [tex]\lim_{n\to\infty} \frac{ \ln (\sqrt{x} +5)}{ \ln \sqrt{x}} = 1[/tex].

    If you want to take your route, you would need to multiply by the log of (sqrtx - 5) instead.
  6. Oct 24, 2007 #5
    if you need to formally show this do you know l'hopital's rule?
  7. Oct 24, 2007 #6
    no, i don't know that rule yet. But gib, the actual problem has ln(x) in the denomator, not the sqr. rt. So does that make it 0?
  8. Oct 24, 2007 #7
    well on a calculator i come up with 1/2
    can someone tell me the best way to go about solving this?
  9. Oct 25, 2007 #8

    Gib Z

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    I know the denominator doesnt have the sqrt mate, but your missing my point. Im saying, The numerator can be replaced with ln(sqrtx) instead of the whole thing, because of the reasons i said before: the plus 5 becomes insignificant as x goes to infinity!! If you just neglected the 5, which you have shown you can do, take mathwonks post into account and take this problem down!!
  10. Oct 25, 2007 #9
    .5 got it thanks
  11. Oct 26, 2007 #10
    Did someone delete my post? ;0
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