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

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.
