Based on the Laplace Transform page in Wikipedia, we can find the function $\mathcal{L}\{f'(t)\}$ by solving $\mathcal{L}\{f(t)\}$ using Integration by Parts:
\begin{align*} \mathcal{L}\{f(t)\} &= \int_{0^-}^{\infty} e^{-st} f(t) \ dt \\ \\ &= \left[\dfrac{f(t)e^{-st}}{-s} \right]_{0^-}^{\infty} - \int_{0^-}^{\infty} \dfrac{e^{-st}}{-s} f'(t) \ dt \\ \\ &= \left[-\dfrac{f(0^{-})}{-s} \right] + \dfrac{1}{s}\mathcal{L}\{f'(t)\} \ , \end{align*}
and isolating $\mathcal{L}\{f'(t)\}$. But on the second line, we evaluate $\left[\dfrac{f(t)e^{-st}}{-s} \right]_{0^-}^{\infty}$ as
\begin{align*} \left[\dfrac{f(t)e^{-st}}{-s} \right]_{0^-}^{\infty} &= \lim_{N\rightarrow\infty} \left[\dfrac{f(t)e^{-st}}{-s} \right]_{0^-}^{N} \\ \\ &= \lim_{N\rightarrow\infty} \left[\dfrac{f(N)e^{-sN}}{-s} - \dfrac{f(0^-)e^{-s(0^-)}}{-s} \right] \\ \\ &= \left[0 - \dfrac{f(0^-)}{-s} \right] \\ \\ &= \left[ - \dfrac{f(0^-)}{-s} \right] \ , \end{align*}
assuming that $s > 0$. My question is, how is it true that $\lim_{N \rightarrow \infty}\dfrac{f(N)e^{-sn}}{-s} = 0$ for all real functions $f(N)$? What if $f(N) = N^N$? Wouldn't the first term diverge if that's the case?