Assume that $f$ is uniformly continuous. Prove that $\lim_{x→∞} f(x) = 0.$ I think the following question is probably fairly easy but can't think of an easy way of proving it.  Some help would be awesome.  This question comes from an old qual.  Thanks.
Let $f$ be an integrable function on $]0, +\infty[.$
(a) Assume that $f$ is uniformly continuous. Prove that $\lim_{x→+∞} f(x) = 0.$
 A: Since $f$ is integrable on $(0,\infty)$ we have that the function $F(x) = \int_{x}^\infty f(t) dt$ is well defined. Moreover, $\lim_{x\to\infty} F(x) = 0$.
Now let $\epsilon > 0$ and by uniform continuity take $\delta > 0$ to be such that $|f(x) - f(y)| < \epsilon$ when $|x-y| < \delta$.
Now consider $F(x) - F(x  + \delta/2) = \int_{x}^{x+\delta/2} f(t) dt = \delta/2 f(\xi_x)$ for some $\xi_x \in [x, x+\delta/2]$ by the mean value theorem for integrals. In particular $|f(\xi_x) - f(x)| < \epsilon$ for all $x$.
Now $F(x + \delta/2) - F(x) \to 0$ as $x \to \infty$, since $F(x) \to 0$. This means $f(\xi_x) \to 0$ as $x \to \infty$. Which means that $f(x)$ is inside $B_\epsilon(0)$ for sufficiently large $x$. Since $\epsilon$ was arbitrary, this means $f(x) \to 0$ as $x\to \infty$.
A: This solution doesn't require the mean value theorem. Suppose the conclusion is false. Then there exists $\epsilon>0$ and an infinite sequence $x_n\in(0,\infty)$ such that $x_n\rightarrow \infty$ and $|f(x_n)|>\epsilon$. We may assume that the $x_n$'s are spaced at least one unit apart. Since $f$ is continuous, $|f(x)|>0$ on some nbd $U_n$ of $x_n$. The $U_n$'s can be assumed to be disjoint. From uniform continuity, $|f(x)-f(t)|<\epsilon/2$ whenever $|x-t|<\delta$. In particular, $|f(x)-f(x_n)|<\epsilon/2$ whenever $|x-x_n|<\delta$ independently of $n$. Take $\delta$ small enough so that $I_n:= (x_n-\delta, x_n+\delta)\subset U_n$. So, $|f(x)|>\epsilon/2$ on $I_n$. Thus,
$\int_0^\infty |f(x)|\,dx \ge \sum_{n=1}^\infty\int\limits_{I_n}|f(x)|\,dx \ge \sum_{n=1}^\infty\epsilon\delta$ which contradicts the integrability of $f$.
