If $f(x^-) < f(x^+)$, what numbers cannot be in the range of $f$? Suppose that $x$ is a number such that $f(x^−) < f(x^+)$ where $f(x^−) := \lim_{t\to x^−}
f(t)$ and $f(x^+) := \lim_{t\to x^+} f(t)$. 
a) What numbers cannot possibly be in the range of $f$? Let’s denote these numbers as $R_x$.
b) If $x$ satisfies $f(x^−) < f(x^+)$, why must there be a rational number in $R_x$?
My solution:
a) $f(x)$
b) Since $f(x^-) < f(x^+)$ and with any two real numbers such that $a < b$, there always exists a rational number between them (I think). Is there a theorem for this?
Looking to see if I have done this correctly and if there is a better solution for b).
 A: The question is easy but since it is fundamental algebra, everything still needs to be proven.
It follows from the definition of a limit that for any sequence $x_k$ monotonically increasing to $x$, the sequence $f(x_k)$ approaches $f(x^-)$. Also, if the function is monotonically increasing, we know that $f(x_{k+1}) > f(x_k)$ for all $k$.
Let $u ≥ f(x^-)$. We prove that $u$ can not be an element of the range of $f$ restricted to $(-\infty, x)$: Let $k \in \mathbb{N}$, then $f(x_k) < f(x^-) ≤ u$. That means $\forall y < x_k, f(y) < u$. Since this holds for any $k \in \mathbb{N}$, we can take the union of all $\{y | y < x_k\}$ which is $\{y | y < x\}$:
$$\forall y < x, f(y) < u.$$
In complete analogy we find that for any $v ≤ f(x^+)$,
$$\forall y > x, f(y) > v.$$
Let now $w$ satisfies both the conditions, that is, $w \in [f(x^-), f(x^+)]$. We know that any $y < x$ and any $y > x$ have functional values which are strictly greater or strictly smaller than $w$.
What about $y = x$? It turns out that the above lines tell nothing of this case. The value of $f(x)$ may coincide with either of the bounds or can lie inbetween (imagine the signum function at zero). Luckily for us, this is just a single value out of a whole interval. So we exclude it and we're done:
$$R_x = [f(x^-), f(x^+)] \setminus \{f(x)\}.$$
Depending on whether $f(x)$ lies strictly between the two bounds, $R_x$ is either a half-open interval or a union of two such intervals. In either case, we use the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$ to prove that a rational number must lie within the set $R_x$.
