Have I correctly proved that $\lim_{||(x,y)||\to\infty}\frac1{y-x}\int_x^y\exp(-1/|t|)dt$ equals $1$? I should prove that, as long as $y\ne x$, $$f(x,y)=\frac1{y-x}\int_x^y\exp(-1/|t|)dt\longrightarrow1 \ \text{as $||(x,y)||\to\infty$}$$ and I would like to do it without $\varepsilon-\delta$ reasoning. My idea was to let $h=y-x$, and then separately consider the cases $|h|\to\infty$ and $|h|\not\to\infty$, as $||(x,y)||\to\infty$.
We have $$f(x,y)=f_h(x)=\frac1{h}\int_x^{x+h}\exp(-1/|t|)dt=\frac1h\int_0^h\exp(-1/|t+x|)dt.$$ If $|h|\to\infty$ then at least one of $|x|$ and $|y|$ must also go to $\infty$. 
In the former case, by the dominated convergence theorem, $$\lim_{||(x,y)||\to\infty}\frac1h\int_0^h\exp(-1/|t+x|)dt=\lim_{||(x,y)||\to\infty}\frac1h\int_0^hdt=1;$$in the latter case, by De L'Hospital, $$\lim_{||(x,y)||\to\infty}\frac1h\int_0^h\exp(-1/|t+x|)dt=\lim_{||(x,y)||\to\infty}\exp(-1/|h+x|)=\lim_{|y|\to\infty}\exp(-1/|y|)=1. $$
Finally, suppose $|h|\not\to\infty$; then $|x|\to\infty$, as a consequence of \begin{align}\sqrt{x^2+y^2}=\sqrt{x^2+(x+h)^2}=\sqrt{2x^2+2hx+h^2}&\le\sqrt{4x^2+4|hx|+h^2} \\ &\le\sqrt{(2|x|+|h|)^2} \\ &=2|x|+|h|.\end{align}So we get once again$$\lim_{||(x,y)||\to\infty}\frac1h\int_0^h\exp(-1/|t+x|)dt=\lim_{||(x,y)||\to\infty}\frac1h\int_0^hdt=\lim_{||(x,y)||\to\infty}\frac{h}h=1.$$
Is my proof correct? Otherwise, how can I amend it?
 A: As your proposed answer stands, you are trying to apply DCT to a two parameter family of functions. But DCT assumes a one parameter family of functions.
However, some of your ideas will work if we're careful. We can generalize to this:
Thm: Suppose $f:\mathbb R\to \mathbb R$ is continuous and $\lim_{|x|\to \infty} f(x) = L\in \mathbb R.$ For $x\ne y,$ define
$$A(x,y) = \frac{1}{y-x}\int_x^y f.$$
Then $\lim_{|(x,y)|\to \infty} A(x,y) = L.$
One special case is easy: $\lim_{|y|\to \infty} A(0,y) = L.$ As you noted, this follows from L'Hopital.
Proof of Thm: First note that the hypotheses on $f$ imply $f$ is bounded, which implies $A(x,y)$ is bounded.
It the theorem fails, then there exists a sequence $(x_n,y_n)$ with $|(x_n,y_n)|\to \infty$ such that $A(x_n,y_n)\to L'\ne L.$ I'll show this leads to a contradiction.
Passing to a subsequence, we can assume WLOG $y_n\to \infty.$ Case i) $(x_n)$ is bounded. Changing variables, we have
$$\tag 1 A(x_n,y_n) = \int_0^1 f(x_n+t(y_n -x_n))\,dt.$$
We can view those integrands as $f_n(t).$ We then have $f_n$ uniformly bounded on $[0,1]$ and $f_n\to L$ pointwise on $ (0,1].$ By the DCT, $(1)\to L,$ contradiction.
Case (ii): $(x_n)$ is unbounded above. Passing to a subsequence, we can assume $x_n\to \infty.$ Then $x_n+t(y_n -x_n)\to \infty$ for each $t\in [0,1].$ Again using DCT, we see $(1)\to L,$ contradiction.
Case (iii): $(x_n)$ is unbounded below. Passing to a subsequence, we can assume $x_n\to -\infty.$ For large $n,$ we have
$$\tag 2 A(x_n,y_n) = \frac{-x_n}{y_n-x_n}A(x_n,0) + \frac{y_n}{y_n-x_n}A(0,y_n).$$
By the special case mentioned above, both $A(-x_n,0),A(0,y_n)$ tend to $L.$ Now use this easily proved result: If both $a_n,b_n\to L,$ and $t_n$ is any sequence in  $[0,1],$ then $t_na_n+(1-t_n)b_n \to L.$ This shows $(2)\to L,$ and again we have a contradiction.
