claims on the basis of $\lim\limits_{x\to\infty}(f*g)(x)=L$ for $f,g ; [0,\infty)\to R$ and $L\in \mathbb R$
assume $\lim\limits_{x\to\infty}(f*g)(x)=L$
Prove the following:
1) $\lim\limits_{x\to\infty}f(x)\lor \lim\limits_{x\to\infty}g(x) $ exist
i know it is right not sure how to explain it
2) if $\lim\limits_{x\to\infty}f(x)$ exist $\Rightarrow \lim\limits_{x\to\infty}g(x) $ exist
not true, i disproof it with example of $f(x)=0$ and $g(x)=D(x)$ (Dirichlet function)
3) if $\lim\limits_{x\to\infty}f(x)=m\;(m\neq0,m\in\mathbb R)$  exist $\Rightarrow\lim\limits_{x\to\infty}g(x) $ exist
i attend to think it is right i am not sure
4) if   $\lim\limits_{x\to\infty}f(x)=\infty$ exist $\Rightarrow \lim\limits_{x\to\infty}g(x) $ exist
cant think of a way to proof or disproof it,
thanks in advance for your time and help!
 A: for 
1) $\lim\limits_{x\to\infty}(f)(x)\lor \lim\limits_{x\to\infty}(g)(x) $ exist
it is not true because we can say that $f(x)=D_1(x)$ and $g(x)=D_2(x)$ Dirichlet functions with opposite values that will give us $\lim\limits_{x\to\infty}(f*g)(x)=0$
A: $1)$ False. Just choose two functions that alternate between $1$ and $0$ and such that when one of them is $1$, the other one is $0$.

$2)$ False. Choose $f(x)=\dfrac1x$ and $g(x)=x$.

$3)$ True. We'll prove $g(x)=\dfrac{L}m$.
Suppose $\displaystyle\lim_{x\to\infty} f\cdot g(x)=L$ and $\displaystyle\lim_{x\to\infty} f(x)=m\neq0$.
Set $\epsilon>0$.
Then there exists an $N_1$ such that for all $x>N_1$, $|f\cdot g(x)-L|<\dfrac{|m|\epsilon}{4}$ and an $N_2$ such that for all $x>N_2$, $|f(x)-m|<\min(\dfrac{|m|}2, \dfrac{m^2\epsilon}{4|L|})$, if $L\neq0$. If $L=0$, then let $N_2=\dfrac{|m|}2$.
Now choose $N=\max(N_1,N_2)$ and suppose $x$ is a number such that $x>N$.
Note $|f(x)-m|<\dfrac{|m|}{2}$ and so $0<\dfrac{|m|}2<|f(x)|$.
Then,
$\left|g(x)-\dfrac{L}{m}\right|=\left|\dfrac{g(x)\cdot f(x)}{f(x)}-\dfrac{L}{m}\right|$ (since $f(x)\neq0$)
Let $a=g(x)\cdot f(x)-L$ and $b=f(x)-m$.
Then the expression becomes, 
\begin{align}
&\left|\dfrac{a+L}{b+m}-\dfrac{L}m\right|
\\\\=&\left|\dfrac{ma+mL-(Lb+Lm)}{(b+m)m}\right|
\\\\=&\left|\dfrac{ma-Lb}{(b+m)m}\right|
\\\\\leq&\dfrac1{|m|}\dfrac{|m||a|}{|b+m|}+\dfrac1{|m|}\dfrac{|L||b|}{|b+m|}
\end{align}
Now, $|b+m|=|f(x)|>\dfrac{|m|}2$.
So,
$\dfrac1{|m|}\dfrac{|m||a|}{|b+m|}+\dfrac1{|m|}\dfrac{|L||b|}{|b+m|}
\\<\dfrac1{|m}\dfrac{2|m||a|}{|m|}+\dfrac1{|m|}\dfrac{2|L||b|}{|m|}
\\=\dfrac{2|a|}{|m|}+\dfrac{2|L||b|}{m^2}
\\<\dfrac{2}{|m|}\dfrac{|m|\epsilon}{4}+\dfrac{2|L|}{m^2}|b|\qquad\left(\because|a|<\dfrac{|m|\epsilon}{4}\right)
\\=\dfrac{\epsilon}{2}+\dfrac{2|L|}{m^2}|b|$
If $L=0$, then the expression equals $\dfrac{\epsilon}2<\epsilon$.
If $L\neq0$, then since $|b|<\dfrac{m^2\epsilon}{4|L|}$, 
$\dfrac{\epsilon}{2}+\dfrac{2|L|}{m^2}|b|<\dfrac{\epsilon}{2}+\dfrac{2|L|}{m^2}\dfrac{m^2\epsilon}{4|L|}=\dfrac{\epsilon}{2}+\dfrac{\epsilon}{2}=\epsilon$.

$4)$ True.
Suppose $\displaystyle\lim_{x\to\infty} f\cdot g(x)=L$ and $\displaystyle\lim_{x\to\infty} f(x)=\infty$.
Then we will prove, $\displaystyle\lim_{x\to\infty} g(x)=0$.
Set $\epsilon>0$.
Then there exists an $N_1$, such that for all $x>N_1$, $|f(x)g(x)-L|<1$.
Also, there exists an $N_2$ such that for all $x>N_2$, $f(x)>\dfrac{|L|+1}{\epsilon}$.
Let $N=\max(N_1,N_2)$, and $x>N$.
Then, $|L|+1>|f(x)g(x)|>\dfrac{|L|+1}{\epsilon}\cdot |g(x)|\implies |g(x)-0|<\epsilon$.
