Let $ f $ be a monotonically increasing function on the interval $ (a, b) $ and let $ x_0 \in (a,b) $ such that $ \displaystyle \lim_{x \to x_0} f(x) = L \in \mathbb{R} $. I need to prove that $ f \ $ is continuous at $ x_0 $. Can you verify my proof?
Suppose, to the contrary, that $ f(x_0) \lt L $. Therefore, $ L - f(x_0) > 0 $, which means that for $ \varepsilon = L - f(x_0) $, there exists a $ \delta > 0 $ such that for all $ x \in (a,b) $, if $ x \in (x_0 - \delta, x_0 + \delta) \setminus \{x_0\} $, then $ -\varepsilon \lt f(x) - L \lt \varepsilon $. Take $ x = \frac{x_0 + x_0 - \delta}{2} $, which is less than $ x_0 $ because $ \delta > 0 $ and is also in $ (x_0 - \delta, x_0 + \delta) \setminus \{x_0\} $. Therefore:
$$ -L + f(x_0) \lt f(x) - L \lt L - f(x_0) \Rightarrow f(x_0) < f(x) $$
However, $ f $ is monotonically increasing, and $ x < x_0 $, which means $ f(x) \lt f(x_0) $ - contradiction.
Suppose, to the contrary, that $ f(x_0) > L $. Therefore, $ f(x_0) - L \gt 0 $, which means that for $ \varepsilon = f(x_0) - L $, there exists a $ \delta > 0 $ such that for all $ x \in (a, b) $, if $ x \in (x_0 - \delta, x_0 + \delta) \setminus \{x_0\}, $ then $ -\varepsilon < f(x) - L < \varepsilon $. Take $ x = \frac{x_0 + x_0 + \delta}{2} $, which is greater than $ x_0 $ because $ \delta > 0 $, and is also in $ (x_0 - \delta, x_0 + \delta) \setminus \{x_0\} $. Therefore:
$$ L - f(x_0) < f(x) - L < f(x_0) - L \Rightarrow f(x) < f(x_0) $$.
However, $ f $ is monotnically increasing, and $ x_0 < x $, which means $ f(x_0) < f(x) $ - contradiction.
Therefore, $ \displaystyle \lim_{x \to x_0} f(x) = L = f(x_0) $, which means $ f $ is continuous at $ x_0 $.