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Simple limit question

  1. Oct 14, 2005 #1
    *Not really homework, just curious:

    [tex] \forall a > 1 , b \ne 0, \lim_{x\to \infty} [ x \ln (a) - \ln (a^x + b) ] = \; {?} [/tex]
     
  2. jcsd
  3. Oct 14, 2005 #2
    Use common logarithm rules to rearrange the expression inside the limit. You should end up with only the limit of a single logarithm.
     
  4. Oct 15, 2005 #3
    I think you mean rearranging the expression as:
    [tex] \mathop {\lim }\limits_{x \to \infty } \ln \left( {\frac{{a^x }}{{a^x + b}}} \right) [/tex]

    Hmm, how would I advance from there? An indeterminate form exists within the natural logarithm, although I'm not quite sure that would justify using L'Hopital.

    Perhaps some additional algebraic manipulation is required?
     
    Last edited: Oct 15, 2005
  5. Oct 15, 2005 #4

    VietDao29

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    Homework Helper

    You can then divide both numerator and denominator within the natural logarithm by ax, something like:
    [tex]\lim_{x \rightarrow \infty} \ln \left( \frac{a ^ x}{a ^ x + b} \right) = \lim_{x \rightarrow \infty} \ln \left( \frac{\frac{a ^ x}{a ^ x}}{\frac{a ^ x + b}{a ^ x}} \right) = \lim_{x \rightarrow \infty} \ln \left( \frac{1}{1 + \frac{b}{a ^ x}} \right)[/tex].
    Can you go from here?
    --------------------
    L'Hopital's rule can be use to solve [itex]\frac{0}{0}[/itex], or [itex]\frac{\infty}{\infty}[/itex] form. It states:
    [tex]\lim_{x \rightarrow \alpha} \frac{f(x)}{g(x)} = \lim_{x \rightarrow \alpha} \frac{f'(x)}{g'(x)}[/tex].
    Note that it can only be used when f(x), and g(x) both tend to 0 or infinity.
    Viet Dao,
     
    Last edited: Oct 15, 2005
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