Let $f:(a,b)\rightarrow \mathbb{R}$ be non-decreasing i.e. $f(x_1)\leq f(x_2)$ and let $c \in (a,b)$. Show that limits from both sides exist. I would like to ask you a question about the following question.
Let $f:(a,b)\rightarrow \mathbb{R}$ be non-decreasing i.e. $f(x_1)\leq f(x_2)$ and let $c \in (a,b)$. Show that $\lim_{x \ \rightarrow \ c-}{f(x)}$ and $\lim_{x \ \rightarrow \ c+}{f(x)}$ both exist.
If $f$ is continuous at $c$ then obviously both limits exist and: $$\lim_{x \ \rightarrow \ c-}{f(x)}=\lim_{x \ \rightarrow \ c+}{f(x)}=f(c)$$ But how would we approach this is $f$ is not continuous at $c$? Thank you guys!
 A: Note that $f(x) \le f(c)$ for $a < x < c$. Therefore the following exists
$$
              \sup_{x < c} f(x)
$$
and you can show that the following limit exists, too:
$$
                  \lim_{x\uparrow c}f(x) = \sup_{x < c} f(x).
$$
A: The function's amplitude on $(a,b)$ is bounded by $m:= f(b)-f(a) \lt \infty$. Look at the right-side limit at a point of discontinuity $c$ (existence of the left-side limit follows in the same way). 
Fix $m \gt \epsilon \gt 0$. Pick a point $x_1$ in the interval to the right of $c$. If $y_1 := f(x_1) - f(c) \lt \epsilon$, you are done (monotonicity). If not, pick a second point $x_2:= \frac{1}{2}(x_1-c)$ in $(c, x_1)$. Calculate $y_2$ as before, and check if it is small enough. Keep this process going until a $y_i \lt \epsilon$ at which point monotonicity finishes the proof. 
If this never happens, we have a sequence $\{x_i\}$ converging from the right to $c$ where all $y_i$ defined as above are greater than $\epsilon$. If the sequence $\{f(x_i)\}$ does not converge, it is not Cauchy. This means that there exists an $\epsilon \gt 0$, such that for each $d \gt 0$ there is a pair $m, n \gt d$ s.t. $$f(x_m) - f(x_n) = |f(x_m) - f(x_n)| \ge \epsilon$$, where we assume wlog $m \lt n$ to fit our construction, and use monotonicity. Note that this is not the same $\epsilon$ as above; it's chosen for notational simplicity. Call this first $n$, $N$. 
But then looking at the sequence only for $k \gt n$, the same applies: we find another pair with indices both larger than $m, n$ whose difference is larger than the same $\epsilon$. Now, for the sum $S$ we have (using 1-4 as indices for simplicity) $$ 2 \epsilon \le S = f(x_1) -f(x_2)+ f(x_ 3)- f(x_4) \le f(x_1) - f(x_4)$$ ($f(x_3) \le f(x_2)$ as $x_3 \le x_2$), so the sum "telescopes" in this way (this is where monotonicity matters again). Iterate this process, and $S$ goes to $\infty$. Hence, $$f(x_N) -f(c) = f(x_N) - f(x_k) + f(x_k) - f(c) \ge f(x_N) - f(x_k) + \epsilon \rightarrow \infty$$ which is a contradiction. 
Hence, the sequence $\{f(x_i)\}$ converges, which was to be shown (this is enough by monotonicity again as all other sequences are sandwiched by this one). 
