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Which grows faster as x-> infinity? ln(x^2+4) or x-5?

  1. Jan 18, 2010 #1
    which grows faster as x--> infinity? ln(x^2+4) or x-5?

    so using L'H rule i got lim as x-->infinity of [ln(x2+4)]/(x-5) = [2x/(x2+4)]/1 = 2x/(x2+4) then using L'H rule again i got 2/2x, then again i 0/2 = 0.

    So, does that mean that x-5 grows faster? And why?
     
  2. jcsd
  3. Jan 18, 2010 #2
    Re: which grows faster as x--> infinity? ln(x^2+4) or x-5?

    You were actually only supposed to (and allowed to) apply L'hopital's rule once. It is obvious that 2x/(x^2 + 4) tends to 0 as x tends to infinity.

    Well, polynomial growth dominates logarithmic growth. This should be clear in even the simplest case where you compare x and log(x).
     
  4. Jan 18, 2010 #3
    Re: which grows faster as x--> infinity? ln(x^2+4) or x-5?

    why are you only allowed to apply L'H rule one time?
     
  5. Jan 18, 2010 #4
    Re: which grows faster as x--> infinity? ln(x^2+4) or x-5?

    Actually in this case it did not matter, and L'Hopital's rule could be applied again. However, if the limit can be computed without L'Hopital's rule, it's sometimes better to do without L'Hopital since you may overlook the hypotheses and try to apply it again incorrectly (this is probably not a big problem on second thought since most people have the 0/0, inf/inf cases ingrained in their heads).
     
  6. Jan 19, 2010 #5
    Re: which grows faster as x--> infinity? ln(x^2+4) or x-5?

    Doesn't the proof for l'Hospital's rule required the limit of f(x)/g(x) to be 0/0 or ∞/∞?
     
  7. Jan 19, 2010 #6

    Matterwave

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    Re: which grows faster as x--> infinity? ln(x^2+4) or x-5?

    x-5 does indeed grow faster than ln(x^2+4). The reason is that an exponential function grows more rapidly than all polynomial functions (as x->inf).

    Don't worry about the constants (-5 and 4) in the expressions, they are of no consequence. So, the comparison is x vs. ln(x^2)

    Now let's compare x vs. ln(e^x) = x. They grow at the same speed. Since e^x grows faster than x^2, then ln(x^2) must grow slower than ln(e^x)=x.

    An intuitive, and non-rigorous proof.
     
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