PTAS vs. FPTAS input
For example, the running time $O(2^{1/\varepsilon} \cdot n^2)$ corresponds to a PTAS, since the running time depends polynomialy on $n$, but not on $1/\varepsilon$ since it is in the exponent. The running time $O(1/\varepsilon \cdot n \log{n})$ depends polynomialy on $n$ and $1/\varepsilon$, so it corresponds to a FPTAS. 

NB, you also need to show that your algorithm has an approximation ratio of $(1 + \varepsilon)$ (or -) to show the algorithm is in fact a (F)PTAS.