Basic analysis question to estimate the growth of a function

Question

Let $f \in C(\mathbb R,\mathbb R)$, $\beta_0 >0$, and $\alpha\in (0,2)$ be given such that \begin{equation}\tag{1}\label{1} \lim_{|s|\to \infty}\frac{|f(s)|}{e^{\beta s^2}} = \begin{cases} 0 & \text{if }\beta > \beta_0 \\ \infty &\text{if } \beta < \beta_0\end{cases} \end{equation} and \begin{equation}\tag{2}\label{2} \lim_{|s|\to 0}\frac{|f(s)|}{|s|^{\alpha/2}} = 0 \end{equation} Now apparently, from \eqref{1} and \eqref{2} we can deduce that there exists some $\beta>0$ such that for each $\varepsilon>0$ there exists a $C_{\varepsilon}>0$ such that $$|f(s)|\leq \varepsilon |s|^{\alpha/2} + C_{\varepsilon}(e^{\beta s^2}-1)\quad \forall s\in \mathbb R$$ I'm not sure how to prove this. Applying the $\varepsilon-\delta$ definitions of the conditions gives me the result near zero and infinity. But how do I "stitch" the results to get it for every $s \in \mathbb R$? Any hints are much appreciated.

Answer 1

Let's say you got the result for $t \in (-\delta, \delta) \cup (-\infty, A) \cup (B, \infty)$ with for example $A \leq -\delta$, $\delta \leq B$ to make the next expression easy to write. Now, the complementary set is $S := [A, -\delta] \cup [\delta, B]$ which is compact, and the function $\displaystyle s \mapsto \frac{|f(s)| - \varepsilon |s|^{\alpha/2}}{e^{\beta s^2} - 1}$ is well-defined and continuous on it, thus that function is bounded on $S$, and you can stitch the results as desired.