Hi I am interested in a result which states that if a continuous function $f: \mathbb{R} \rightarrow \mathbb{R}$ has finite limits on both sides then the function is uniformly continuous. Is it possible to apply this somehow to state that $e^{-x}$ is uniformly continuous on $(0,\infty)$ by noting that $\lim\limits_{x \rightarrow 0}e^{-x} = 1$ and $\lim\limits_{x \rightarrow \infty} e^{-x} = 0$, both of which are finite.
I know how to prove this by other means, for example using the Mean Value Theorem but I am interested in this way. Thanks.
Let $\epsilon > 0$. Since $f$ has finite limits on both sides, call them $n$ and $p$, there are $N, P$ so that for $x< N$ an $x>P$ you have $|n - f(x)| \le \epsilon$ and $|p - f(x)| \le \epsilon$, respectively. Thus for $x,y < N$ you have $|f(x)- f(y)| \le 2 \epsilon$ and the same on the positive side. So for large and small $x$. There is no issue.
Now, consider the function restricted to $[N-1, P+1]$, say. This is a compact interval. So continuity implies uniform continuity on that interval.
You can then put these two (or three) things together to get uniform continuity.
You can argue in a similar way if the limits are not towards infinity but some other value. The key is always. Treat a neighborhood of the limit point in one way, and for the rest use compactness.