Decay at infinity on time and space for the heat equation solution

Question

I'm trying to solve the following exercise: let $u(x,t)$ be the classic solution to $u_t=\alpha u_{xx}$ on $\mathbb{R}\times (0,\infty)$, with initial condition $u(x,0) = f(x)\in\mathcal{S}(\mathbb{R})$. Then I must show that $$\lim_{|x|+t\to \infty}u(x,t) = 0$$ I've managed to reduce the problem to proving that $$\frac{1}{\sqrt{4\pi\alpha t}}\int_{-R}^Re^{\frac{-(x-y)^2}{4\alpha t}}dy\to0\text{ as }|x|+t\to\infty$$ for some $R>0$. The problem I'm having is dealing with this type of limit in both $x$ and $t$ at the same time, since I can't guarantee that $t$ doesn't tend to $0$ or $x $ stays bounded... I've tried to use the limited convergence theorem, but I'm not sure how to justify this joint limit. Does $$\frac{1}{\sqrt{4\pi\alpha t}}e^{\frac{-(x-y)^2}{4\alpha t}}\to0\text{ as }|x|+t\to\infty$$ for each $y$? How can I prove it? If so, if I could limit $$\frac{1}{\sqrt{4\pi\alpha t}}e^{\frac{-(x-y)^2}{4\alpha t}}$$ by some constant then I would be done, but I'm not sure how to limit it. Alternatively, the book suggests bounding $$|u(x,t)|\leq \frac{C}{1+|x|^2}+Ct^{-1/2}e^{-cx^2/t}$$ for when $|x|>t$, but I don't see how to do it...