Using big Oh notation.

Question

How to show that $O(x^{1/2} e^{ (log (\frac{x}{c}))^{1/2}})=O(x^{1/2+\epsilon})$ for any $\epsilon>0$?

Answer 1

Since $$ y - k = O(\varepsilon^2 y^2) $$ one has that for all $\varepsilon$ and $c$ $$ \log \frac{x}{c} = \log x - \log c= O(\varepsilon^2\log^2 x) $$ which means that $$ (\log\frac{x}{c})^{\frac 1 2} = O(\varepsilon \log x) $$ and hence $$ e^{(\log\frac{x}{c})^{\frac 1 2}} = O(e^{\varepsilon \log x })= O(x^\varepsilon). $$ The required result follows.

Answer 2

Let $f(x)=e^{(\log(x/c))^{1/2}}=e^{\sqrt{\log(x/c)}}$
Then we have $f(x)^{\sqrt{\log(x/c)}}=e^{(\log(x/c))}=x/c$
So we can see that $f(x)^{\sqrt{\log(x/c)}}=x/c$ which means $f(x)=(x/c)^{\frac{1}{\sqrt{\log(x/c)}}}<x^{\epsilon}$ since $\frac{1}{\sqrt{\log(x/c)}}\rightarrow 0$.

Combining all together yields $x^{1/2}e^{(\log(x/c))^{1/2}}<x^{1/2+\epsilon}$
which gives immediately what yu want.