$\lim\limits_{x \to a}f(x)$ and $\lim\limits_{x \to a}g(x)$ do not exist, $\lim\limits_{x \to a}[f(x)+g(x)]$ exists, but $g \neq -f$?

Question

I am tutoring calculus to a friend, and he managed to find an example of two functions $f$, $g$ such that $\lim\limits_{x \to a} f(x)$ and $\lim\limits_{x \to a} g(x)$ do not exist, but $\lim\limits_{x \to a} [f(x) + g(x)]$ exists. As we discussed making this more general, we noticed that if we took a function $f$ such that $\lim\limits_{x \to a}f(x)$ does not exist, the function $g = -f$ (assuming $\lim\limits_{x \to a}g(x)$ does not exist) would satisfy the condition that $\lim\limits_{x \to a} [f(x) + g(x)]$ exists.

To try to extend this even more, we wondered if there were two functions $f$, $g$ such that at some point $a \in \mathbb{R}$, all of the following are satisfied:

1. $\lim\limits_{x \to a} f(x)$ does not exist
2. $\lim\limits_{x \to a} g(x)$ does not exist
3. $\lim\limits_{x \to a} [f(x) + g(x)]$ exists
4. $g \neq -f$ in $\mathbb{R}$

Since we haven't discussed infinite limits yet, assume $\left|\lim\limits_{x \to a} [f(x) + g(x)]\right| < \infty$. Are there any examples of $f$, $g$ that satisfy the above four conditions?

As you might guess, none of us were sure where to even begin with this, so unfortunately, I don't have any more thoughts other than the above. I would prefer an answer which would be understandable to someone new to calculus, but if it requires some $\delta$-$\epsilon$, I can try to teach him this as well. The more elementary and motivated the explanation, the better.

Answer 1

Yes, there are. The simplest ones (like the one below) might feel a bit like cheating, though, so be warned (I've chosen $a = 0$). $$ f(x) = \frac1x, \quad g(x) = x-\frac1x $$ In essence, recognizing that $f+g$ is more or less well-behaved at $x = a$ (or, at least, we allow it to be), but (4) says, right out, that it's not the $0$ function, we just choose some nice function $h(x)$ which is not $0$ everywhere, choose an "ugly" $f(x)$, and make $g(x) = h(x) - f(x)$.

Answer 2

You can take $$f (x)=\sin^2 (\frac {1}{x} )$$ $$g (x)=\cos^2 (\frac {1}{x}) $$ $$a=0$$

$$\lim_0 (f (x)+g (x))=1$$ but $\lim_0 f (x) $ and $\lim_0g (x) $ do not exist.