In the complexity analysis of an algorithm, I encountered this function in $x$ for some small positive constants $a$, $b$ with $a>b>1$.
$$f(x)={(a/b)^{\log_a(x(a-1)+1)}-1\over x(a/b-1)}$$
What is the asymptoptic growth of $f(x)$ as $x\to\infty$? Note that $f(x)\to0$ as $x\to\infty$ and I believe there are asymptoptics of the form
$$f(x)\sim{c\over\root d\of x}+\mathcal{O}(x^{-1})$$
for some constants $c$, and $d$ but I didn't find any way to isolate $x$.
After a great deal of fiddling with the expression, I get
$$\def\O{\mathcal{O}}\eqalign{f(x)&={(a/b)^{\log_a(x(a-1)+1)}-1\over x(a/b-1)}\cr &={b\over x(a-b)}\Biggl(\left({a\over b}\right)^{\log_a(x(a-1)+1)}-1\Biggr)\cr &={b\over x(a-b)}\Biggl(\left({a\over b}\right)^{\log_ax+\log_a(a-1)+\O(x^{-1})}-1\Biggr)\cr &\approx{b\over x(a-b)}\left(x^{\log_a(a/b)}\,(a-1)^{\log_a(a/b)}-1\right)\cr &={b\over x(a-b)}\left(x^{1-1/\log_ba}\,(a-1)^{1-1/\log_ba}-1\right)\cr &={b\over a-b}\left({{a-1\over\root\log_ba\of{x(a-1)}}}-{1\over x}\right)=\O\left(x^{-1/\log_ba}\right).}$$