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Horizontal asymptote

  1. Oct 30, 2005 #1
    is there a horizontal asymptote for

    i know u take the lim to find h.a
    but what is ln infinity/infinity?
  2. jcsd
  3. Oct 30, 2005 #2
    I could be wrong, but I think L'Hopitals Rule could work here.
  4. Oct 30, 2005 #3
    what is L'Hopitals Rule ?
  5. Oct 30, 2005 #4
    Because your limit produces an indeterminate form (that infinity/infinity), you can differentiate separately the numerator and denominator with respect to [itex]x[/itex], thus applying L'Hopital's Rule :biggrin:. You see :shy:,

    [tex] \mathop {\lim }\limits_{x \to \infty } \frac{{\ln x^2 }}{{x^2 }} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{d}{{dx}}\left( {\ln x^2 } \right)}}{{\frac{d}{{dx}}\left( {x^2 } \right)}} = \mathop {\lim }\limits_{x \to \infty } \frac{1}{{x^2 }} = 0 [/tex]

    Hope this helps :smile:
  6. Oct 30, 2005 #5
    o..icic...thats the same for -infinity right?
    so the horizontal asymptote is y=0 ?
  7. Oct 30, 2005 #6
    Correct .
    Last edited: Oct 30, 2005
  8. Oct 30, 2005 #7
    i see...thanks a lot
  9. Oct 30, 2005 #8
    Another, quicker way to do this would be to recognize the functions you are dealing with. An algebraic function will "dominate" a logarithmic function for large values of x. Knowing this means that the denominator will go to infinity quicker than the numerator, thus going to zero.
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