$f$ uniformly differentiable function, $\int_{0}^{\infty}f$ converges, and $f'$ is bounded. Prove: $\lim f(x)=0$ when $x \to \infty$. $f:\mathbb{R}\to \mathbb{R} $ uniformly d,ifferentiable function, $\int_{0}^{\infty}f$ converges, and $f'$ is bounded. I need to prove that $\lim f(x)=0$ when $x \to \infty$.
This is what I did so far:
I chose a series $c_n \to \infty$ such that $f(c_n)> \epsilon$, assuming that is$\lim f(x)\neq0$ when $x \to \infty$. then I wrote $f(c_n+t)=f(c_n)+ \int_{c_n}^{c_n+t}f'(x) \geq f(c_n)-tM \geq\epsilon-tM$. I don't really know if this is useful, I can't see the whole picture. I'd love to know if this is useful and to read a possible solutions.
Thanks!
 A: If $\lim_{x\to +\infty}f(x)\neq 0$, then, substituting $f$ by $-f$  is necessary, we can find a sequence $\{x_n\}$ with converges to $+\infty$ and $\delta_0>0$ such that $f(x_n)\geq 2\delta_0$. Put $M:=1+\sup_{x\geq 0}|f'(x)|$. We can extract from $\{x_n\}$ a subsequence, which will be denoted $\{t_n\}$ such that $t_{n+1}-t_n\geq\frac{2\delta_0}M$. We have $|f(t_n)-f(x)|\leq M|t_n-x|$ so if $|t_n-x|\leq \frac{\delta_0}M$ we have $f(x)\geq \delta_0$. Indeed, $|f(t_n)-f(x)|\leq \delta_0$ so $2\delta_0-f(x)\leq f(t_n)-f(x)\leq \delta_0$, and $\delta_0-f(x)\leq 0$.
Hence $$\forall n\in\mathbb N:\int_{t_n-\frac{\delta_0}M}^{t_n+\frac{\delta_0}M}f(t)dt\geq2\frac{\delta_0^2}M,$$
but this contradicts the fact that the integral $\int_0^{+\infty}f(t)dt$ is convergent. Indeed, if it was the case we would be able to find $c$ such that if $x,y\geq c$ then $\left|\int_x^yf(t)dt\right|\leq \frac{\delta_0^2}M$.

In fact, it can be shown by a similar argument that an uniformly continuous function on $[0,+\infty[$ such that the integral $\int_0^{+\infty}f(t)dt$ is convergent has a limit $0$. The arguments are quite the same, since we get $\eta$ such that if $|x-y|\leq \eta$ then $|f(x)-f(y)|\leq \delta_0$.
