If $f$ has at most polynomial growth, then $g(x)=\int_0^\infty f(x,t)e^{-t}\;dt$ is continuous I have the following question from a past analysis qualifying exam:

Let $f$ be a continuous real-valued function with two variables and at most polynomial growth. That is, there exists a constant $N$ such that $|f(x,t)|\le N(1+x^2+t^2)^N$ for all $(x,t)\in\mathbb{R}^2$. Consider the function $g$ defined by
  $$g(x)=\int_0^\infty f(x,t)e^{-t}\;dt.$$ Prove that $g$ is a continuous function on the real line $\mathbb{R}$.

Let $\epsilon>0$ and let $x,y\in\mathbb{R}$. Then
$$|g(x)-g(y)|=\left|\int_0^\infty f(x,t)e^{-t}\;dt-\int_0^\infty f(y,t)\;dt\right|=\left|\int_0^\infty(f(x,t)-f(y,t))e^{-t}\;dt\right|$$
$$\le\int_0^\infty|f(x,t)-f(y,t)|e^{-t}\;dt$$
I am having trouble pulling a factor of $|x-y|$ out here so that I can choose a $\delta$ accordingly. At first I wanted to apply the triangle inequality so that I could use the assumption that $|f(x,t)|\le N(1+x^2+t^2)^N$, but this doesn't seem like the way to go. Any help would be much appreciated.
 A: Here's a sketch of how one might do this: fix $\epsilon>0$ and take a given $x$. To show continuity, we must find a $\delta>0$ such that $\vert g(y)-g(x)\vert<\epsilon$ when $\vert y-x\vert\leq \delta$. First, use the polynomial growth condition on $f$ to show that there exists some constant $K$ (depending only on $x$ and $\epsilon$) such that for all $z\in [0,x+1/2]$, say, $\int_K^{\infty}\vert f(z,t)e^{-t}\vert dt<\epsilon/4$ (this follows by integrable tails and the growth condition). 
Next, note that on the box $[x-1/2,x+1/2]\times [0,K]$, $f(x,t)$ is uniformly continuous as a continuous function on a compact set. In particular, this implies that there exists some $\delta'>0$ such that for $\vert x-y\vert< \delta'$ inside this box, we have $\vert f(x,t)-f(y,t)\vert<\frac{\epsilon}{2K}$.
Finally, take $\delta=\min\{1/2,\delta'\}$. Then with $\vert x-y\vert< \delta$, we can split up the computation as you did:
\begin{align}
\vert g(x)-g(y)\vert&\leq \int_0^{\infty} \vert f(x,t)-f(y,t)\vert e^{-t}dt\\
&=\int_0^K \vert f(x,t)-f(y,t)\vert e^{-t}dt+\int_K^{\infty} \vert f(x,t)-f(y,t)\vert e^{-t}dt\\
&\leq \frac{K\epsilon}{2K}+2\frac{\epsilon}{4}\\
&=\epsilon,
\end{align}
where we use the triangle inequality and the bounds found above on the second inequality.
