If $\lim f(x)$ and $\lim g(x)$ do not exist, can the $\lim [f(x)+g(x)]$ exist? 
This is a question from Calculus by Michael Spivak, how do start answering this question?
 A: Yes the addition can definitely exist, just take the following two step functions:
$$
f(x) = \begin{cases}
1 & x \geq 0 \\
-1 & x < 0 \end{cases} \\
g(x) = \begin{cases}
-1 & x \geq 0 \\
1 & x < 0 \end{cases} \\
f(x) + g(x) = \begin{cases}
0 & x \geq 0 \\
0 & x < 0 \end{cases} \rightarrow f(x) + g(x) = 0
$$
This is how you need to think.  The same thing can happen when multiplying.  Perhaps the first function is $0$ on the left and the second function is $0$ on the right.  Then at the point of discontinuity they could both multiply (on each side) to give $0$ from both sides.
Note that you don't always have to make the resulting function equal to $0$ (just look at the above step functions, I could just shift both functions up by $1$ and then the addition would be $f(x) +g(x) = 2$ not $0$).
A: $f(x)=\dfrac1{x-a}$ and $g(x)=\dfrac1{a-x}$
A: a) Yes addition can exists if $g = -f$, we have $\lim_ag$ does not exist but $\lim_a f+g=\lim_a0=0$ exists. If $f(x)\not=0$, then $\lim_ag=\lim_a1/f$ does not exists but $\lim_a fg=\lim_a1=1$ exists. 
b) If $\lim_a f$ and $\lim_a f+g$ exists then $\lim_ag=\lim_a[(f+g)-f]=\lim_a(f+g)-\lim_af$. Thus must exists.
c) Suppose for sake of contradiction that $\lim_a f+g$ exists.  Then by (b) it follows that $\lim_a g$ must exists. Contradiction.
d) No. If $\lim_af=0$ then we cannot use an argument as (a). For example suppose that $g(x)=-1$ for $x\le a$ and $g(x)=1$ for $x>a$. Then 
$$\lim_{\uparrow a}fg=\lim_{\uparrow a}-f=0;\;\;\lim_{\downarrow a}fg=\lim_{\downarrow a}f=0$$
Thus $\lim_a fg=0$. But $\lim_a g$ does not exist.
A: For (a), let $f =1 + 1_\mathbb{Q}$, $g=-f$. Then $\lim_{x \to a} f(x)$ does not exist for any $a$ and similarly for $g$.
Note that $f+g = 0$, hence the limit of the sum exists, and note that ${f \over g} = -1$, hence the limit of the quotient exists.
For (b) since $g(x) = (f(x)+g(x)) - f(x)$ and subtraction is continuous, we see that $\lim_x g(x) = \lim_x (f(x)+g(x)) - \lim_x f(x)$.
(c) follows immediately from (b).
For (d), let $f(x) = 0$, and $g$ be the $g$ from Part (a). Then we see that $\lim_x (f) = 0, \lim_x f(x)g(x) = 0$, but clearly $\lim_x g(x) $ still does not exist.
A: take $f$ be a function limit not existing on $a$
and let $g=-f$
thus when x goes to $a$ limit not exists
but $f+g=f-f=0$ 
thus limit of $f+g=f-f$ is exists and equal to zero
