If $\lim_{x\to0} f(x)+g(x)$ and $\lim_{x\to0} f(x)g(x)$ exist simultaneously, are there any $f(x)$ and $g(x)$ that do not have limit If $\lim_{x\to0} f(x)+g(x)$ and $\lim_{x\to0} f(x)g(x)$ exist simultaneously, are there any $f(x)$ and $g(x)$ that do not have their own limits?
 A: One can even find an example where the functions $f$ and $g$ are nowhere continuous: take $f(x)=\begin{cases}1,&x\in\Bbb Q,\\0,&x\notin \Bbb Q,\end{cases}$ and $g(x)=1-f(x)$.
Obviously, $\forall x$ $f(x)+g(x)=1$, $f(x)g(x)=0$.
A: There exist examples like $f = \left\{ \begin{array}{ll}0 & x \ge 0\\ 1 & x < 0 \end{array}\right., 
g = \left\{ \begin{array}{ll}1 & x \ge 0\\ 0 & x < 0 \end{array}\right.$.  Here $f(x) + g(x)$ is identically 1 and $f(x)g(x)$ is identically zero.  However, neither function has a limit at $x=0$.
A: This answer is intended to be a general understanding for the question.
Set 
$$
\mathop{\mbox{sgn}}\big(g(x)-f(x)\big)=
\begin{cases}
\dfrac{g(x)-f(x)}{|g(x)-f(x)|} & \mbox{if } \;\;\;\ g(x)-f(x)\neq 0\\
\\
 \;\;0 & \mbox{if } \;\;\;g(x)-f(x)=0\\
\end{cases}
$$ 

Proposition. Supose $\lim_{x\to 0} [g(x)+f(x)]=S$ and $\lim_{x\to 0} [g(x)\cdot f(x)]=P$. If there is the limit $\lim_{x\to 0} \mathop{\mbox{sgn}}\big(g(x)-f(x)\big)$ then there are limits 
  $$
\lim_{x\to 0}g(x)= S+\lim_{x\to 0}\mathop{\mbox{sgn}}\big(g(x)-f(x)\big)\cdot\sqrt{S^2-4P}
\\
\lim_{x\to 0}f(x)= S-\lim_{x\to 0}\mathop{\mbox{sgn}}\big(g(x)-f(x)\big)\cdot\sqrt{S^2-4P}
$$
  If there is not the limit $\lim_{x\to 0} \mathop{\mbox{sgn}}\big(g(x)-f(x)\big)$ then there no are limits $\lim_{x\to 0}g(x)$ and $\lim_{x\to 0}f(x)$.

Proof. Note that, 
$$
\big|g(x)-f(x)\big|= \sqrt{[g(x)+f(x)]^2-4[g(x)\cdot f(x)]}=\sqrt{g(x)^2-2g(x)\cdot f(x)+f(x)^2}
$$
and 
$$
\big[ g(x)-f(x) \big]=\mathop{\mbox{sgn}}\big(g(x)-f(x)\big)\cdot\big|g(x)-f(x)\big|.
$$
Therefore we can conclude that
\begin{align}
g(x)=&\frac{1}{2}\big[ g(x)+f(x) \big]+
\frac{1}{2}\mathop{\mbox{sgn}}\big(g(x)-f(x)\big)
\cdot\left|g(x)-f(x)\right| \\
f(x)=&\frac{1}{2}\big[ g(x)+f(x) \big]
-\frac{1}{2}\mathop{\mbox{sgn}}\big(g(x)-f(x)\big)
\cdot\big|g(x)-f(x)\big|\\
\end{align}
Observe that $\lim_{x\to 0}|f(x)-g(x)|=\sqrt{S^2-4P}$.
 If $\lim_{x\to 0}
\mathop{\mbox{sgn}}\big(g(x)-f(x)\big)$ exist ( or not), then follows the result.
A: Assume that the functions $m$ and $p$ are defined in a punctured neighborhood $\dot U$ of $x=0$ and that the limits $\lim_{x\to0} m(x)$ and $\lim_{x\to0}p(x)$ exist.  We are asking for functions $f$ and $g$ such that
$${f(x)+g(x)\over2}=m(x),\qquad f(x)\>g(x)=p(x)\ .\tag{1}$$
When $m^2(x)-p(x)\geq0$ in $\dot U$ we can define the functions
$$h_+(x):=m(x)+\sqrt{m^2(x)-p(x)},\quad  h_-(x):=m(x)-\sqrt{m^2(x)-p(x)}\qquad(x\in\dot U)\ .$$
Together they solve $(1)$ in place of $f$ and $g$, and from the continuity of the involved operations it follows that the limits $\lim_{x\to0}h_+(x)$, $\>\lim_{x\to0}h_-(x)$ exist.
But there are infinitely many solutions $f$, $g$ of $(1)$ for whom the corresponding limits do not exist. The most general solution of $(1)$ is given by
$$f(x):=m(x)+\sigma(x)\sqrt{m^2(x)-p(x)},\qquad  g(x):=m(x)-\sigma(x)\sqrt{m^2(x)-p(x)}\ ,$$
where $$\sigma: \quad \dot U\to\{-1,1\},\qquad x\mapsto\sigma(x)$$ is  arbitrary. 
