Limit as $x$ goes to infinity What is the limit of $\,x^a\big/2^{\log x}$ as $x$ goes to infinity?  L'Hopital's rule doesn't help and I can't think of anything that can be done algebraically to make this expression more manageable.  

Because $u^v = \exp(v\log u)$, the expression is
$$ \exp(a\log x- \log x \cdot \log 2 ) = \exp(\log x\cdot (a-\log 2)) = x^{a-\log 2} $$
If the exponent is positive, i.e. if $a\gt \log 2$, the limit $x\to\infty$ is $\infty$. If $a=\log 2$, the limit is one because the function is one everywhere. If $a\lt \log 2$, the limit is $0$.
