if the improper integral $\int^\infty_a f(x)\,dx$ converges, then $\lim_{x→∞}f(x)=0$ I need to prove that:
$$\lim_{x→∞}f(x)=0$$
if 
$$\displaystyle∫^∞_af(x)\,dx$$ converges.
I need a proof or an specific, and if possible simple, counterexample. Would really appreciate your help! Thank you in advance.
 A: This is false; here is a counterexample:
$$
\begin{align}
\int_0^\infty \sin(x^2)\, dx
&=\int_0^\infty \frac{\sin x}{2\sqrt x}\,dx\\
&= \frac{1}{2} \sum_{n=0}^\infty \int_{n\pi}^{(n+1)\pi} \frac{\sin x}{\sqrt x}\,dx\\
&= \frac{1}{2} \sum_{n=0}^\infty (-1)^n \left|\int_{n\pi}^{(n+1)\pi} \frac{\sin x}{\sqrt x}\,dx\right|
\end{align}
$$
converges by the alternating series test (in fact, it equals $\sqrt{\frac{\pi}{8}}$).
However,
$$\lim_{x\to\infty}\sin(x^2)$$
does not exist.  
A: The function
$$
\int_0^\infty|\sin(x)|^{2\left\lfloor x/\pi\right\rfloor^3}\,\mathrm{d}x\tag{1}
$$
converges, but $|\sin(x)|^{2\left\lfloor x/\pi\right\rfloor^3}$ does not converge to $0$.
Note that
$$
\begin{align}
\int_{n\pi}^{(n+1)\pi}|\sin(x)|^{2\left\lfloor x/\pi\right\rfloor^3}\,\mathrm{d}x
&=\int_{0}^{\pi}|\sin(x)|^{2n^3}\,\mathrm{d}x\\
&=\frac\pi{4^{n^3}}\binom{2n^3}{n^3}\\
&\le\sqrt{\pi/n^3}\tag{2}
\end{align}
$$
and so $(1)$ converges by comparison to
$$
\pi+\sum_{n=1}^\infty\sqrt{\pi/n^3}\tag{3}
$$
A: $\lim_{x\to\infty}F(x) - F(a)$ converges implies that $x\to\infty \Longrightarrow F(x) \to F(a)$. But for a sequence to be convergent, it has to converge to one specific value - meaning, $F(a)$ and $F(x)$ converge to this one unique value for convergence.

 HINT: ZERO.

A: If you dont need it to be continuous thake $f(x)= 1_{\mathbb{N}(x)}$ the characteristic function
of $\mathbb{N}$.
Otherwise just make this function continuous making sure the $\int_{0}^{\infty}f(x)\rm{d}x$
converges. 
How you can do that? Imagine the graph of $f(x)$ to be triangles where one vertex will be 
$f(n)$ and the other two and $x-$axis. Let these two points be $x_1,x_2$ now the area of that triangle will be $(x_2-x_1) \frac{1}{2}$ so by picking $x_1,x_2$ close enough you can ensure that the integral converges. And by construction of course you get $\lim f(x) \ne0$   .
A way to write it down explicitly is the following
$$
f(x)=
\left \{
\begin{array}{cc}
-|x|+1& x\in(-1,1)\\
0&\rm{otherwise}
\end{array}
\right.
$$
Define $g(x)= \sum_{n=0}^{\infty}f\big ( (x-n)2^n\big) $
A: This is false, but constructing a counterexample is kind of tricky. You might want to start by thinking about $g(x) = \int_a^x f(s)\, ds-\int_a^\infty f(s)\, ds$ instead of $f$. You are then looking for a $g$ with $\lim_{x\to\infty} g = 0$ but $\lim_{x\to\infty} g'$ does not exist. Think about what this would mean: $g$'s amplitude has to die down the larger $x$ gets, but its slope (i.e., frequency) has to change more and more wildly.
A: Let $f(x)=\sum_{n=1}^{\infty}f_n(x)$ where $f_n(x)=n\cos[\frac{n^3\pi}2(x-n)]$ when $|x-n|\le\frac1{n^3}$ and $f_n(x)=0$ otherwise.
