Limit of x ln(a) - ln(a^x+b) for a>1, b!=0

[bomba923]

*Not really homework, just curious:

\forall a > 1 , b \ne 0, \lim_{x\to \infty} [ x \ln (a) - \ln (a^x + b) ] = \; {?}

 

[hypermorphism]

Use common logarithm rules to rearrange the expression inside the limit. You should end up with only the limit of a single logarithm.

 

[bomba923]

I think you mean rearranging the expression as:
\mathop {\lim }\limits_{x \to \infty } \ln \left( {\frac{{a^x }}{{a^x + b}}} \right)

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?

 

[VietDao29]

You can then divide both numerator and denominator within the natural logarithm by a^x, something like:
\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).
Can you go from here?
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L'Hopital's rule can be use to solve \frac{0}{0}, or \frac{\infty}{\infty} form. It states:
\lim_{x \rightarrow \alpha} \frac{f(x)}{g(x)} = \lim_{x \rightarrow \alpha} \frac{f'(x)}{g'(x)}.
Note that it can only be used when f(x), and g(x) both tend to 0 or infinity.
Viet Dao,