A continuous $f:\RR\to\RR$ vanishing at $\pm\infty$ is uniformly continuous

Fix \(\eps>0\), we need to find a \(\delta = \delta(\eps)\) such that \[ \abs{x-y}<\delta \implies \abs{f(x) - f(y)} < \eps && \forall x, y\in \RR .\] Use that \(\lim_x\to \pm \infty f(x) = 0\) to choose \(M\gg 0\) such that \[ \abs{x} \geq M \implies \abs{f(x)} \leq \eps/2 ,\] then \[ \abs{x}, \abs{y} \geq M \implies \abs{f(x) - f(y)} \leq \abs{f(x)} + \abs{f(y)} \leq \eps .\] So in this region choose (say) \(\delta_1 < \eps\) to ensure that \(B_\delta(x), B_\delta(y) \subseteq [-M, M]^c\). On \([-M, M]\), note that this region is compact and \(f\) continuous on a compact set implies uniformly continuous. So use this to choose \(\delta_2 = \delta_2(\eps)\) in this region to ensure \(\abs{f(x) - f(y)} < \eps\).

This handles the cases \(x, y \in (M, M)^c\), or \(x,y\in [M, M]\), so it only remains to handle \(x\in [M, M]\) and \(y\in (M, M)^c\) (wlog, relabeling \(x,y\) if necessary). In this case, use the triangle inequality: \[ \abs{f(x) - f(y)} &= \abs{f(x) - f(M) + f(M) -f(y)} \\ &\leq \abs{f(x) - f(M)} + \abs{f(M) -f(y)} \\ &\leq \eps + \abs{f(M)} + \abs{f(y)} \\ &\leq \eps + \eps + \eps ,\] where we’ve used that \(M, y\in (M, M)^c\) to apply the first bound and \(M, x\in [M, M]\) to apply the second.