Given $f :(a,b)\to\mathbb{R}$ is a monotone increasing function bounded above, show that $\lim_{x\to b^-} f(x)$ exists Let $f :(a,b)\to\mathbb{R}$ be a monotone increasing function bounded above. Show that $\lim_{x\to b^-} f(x)$ exists. 
Okay, so what I've done is shown below. Apparently it's not enough because I've just shown that $f(x_n)$ converges. Could anyone give any hints as to how I can complete the proof?
Let $x_n=b-\dfrac{1}{n},n\in\mathbb{N}$. Then since $\forall n\in\mathbb{N},x_n<b$ and $x_{n+1}=b-\dfrac{1}{n+1}>b-\dfrac{1}{n}=x_n$, $x_n$ is monotone increasing and bounded above. Thus, by the Monotone Convergence Theorem, it has a limit $\lim_{n\to\infty}x_n=b^-$. Assume that $f :(a,b)\to \mathbb{R}$ is monotone increasing and bounded above by some $L\in\mathbb{R}$. Since $f(x)$ is monotone increasing, we know that $f(x_n)<L \forall n\in\mathbb{N}$. But here I'm stuck. How do I show precisely that the given limit exists?
 A: Application of some sequence. 
Let $x_n = b -1/n$. Then $x _n \to b^-$. Since $f$ is increasing and bounded above, so is the sequence $(f(x_n))_1^\infty$. Therefore  it converges. Let the limit be $A$. By definition, for each $\varepsilon > 0$, there is some $N \in \Bbb N^*$ s.t. $\vert f(x_n) - A\vert < \varepsilon$ whenever $n \geqslant N$. Therefore for $\underline {\delta = b - x_N}$, whenever $\underline{x_N = b -\delta < x < b}$, there is an $M \in \Bbb N^*$ that $M \geqslant N$ and $x_M \leqslant x \leqslant x_{M+1}$. Then by monotonicity, $f(x_M) \leqslant f(x) \leqslant f(x_{M+1})$, and 
$$
\underline {- \varepsilon <}\ f(x_M) - A \leqslant \underline {f(x) - A} \leqslant f(x_{M+1}) - A\ \underline { < \varepsilon}\ , 
$$
i.e.
$$
\lim_{ x \to b^-} f(x) = A. 
$$
A: Since $f$ is bounded above, the set $f(a,b)=\{f(c):a<c<b\}$ must have a least upper bound $u$. We will show that $\lim_{x\to b^-}f(x)=u$. 
Take any $\varepsilon>0$. Since $u$ is a least upper bound, we have that 
$u-\varepsilon$ is no longer an upper bound, and hence there is some $c<b$ with 
$f(c)>u-\varepsilon$. Let $\delta=b-c$. Then if $x<b$ and $b-x<\delta$ we have that 
$c<x<b$ and by monotonicity $u-\varepsilon<f(c)\le f(x)\le u$. 
Edit. Here is an attempt to incorporate/edit what the OP had, together with a hint by @xbh, and try to use sequences. So the sequence $x_n=b-\frac1n$ is monotone increasing (not decreasing) and converges to $b$ (since $\frac1n\to0$). The sequence $f(x_n)$ is also monotone increasing (since $f$ is monotone increasing, and $x_n$ are monotone increasing) and since $f(x_n)$ is bounded above, it must have a limit $L$, that is $f(x_n)\to L$, monotonically increasing, as $n\to\infty$. (We did just use the Monotone Convergence Theorem.) 
It remains to show that $\lim_{x\to b^-}f(x)=L$. Take any $\varepsilon>0$. There is $N$ big enough such that $f(x_n)>L-\varepsilon$ whenever $n>N$. 
Let $\delta=\frac1{N+1}$. At this point I do not see what is the virtue of using a sequence, instead of $\varepsilon-\delta$ definition of limit, at any rate take $x$ with $b-x<\delta$, so then $x>b-\delta=x_{N+1}$, so $L-\varepsilon<f(x_{N+1})\le f(x)\le L$. 
One may wish to involve sequences even more, and prove that for every sequence $y_k\to b^-$ (not necessarily increasing) we have that $f(y_k)\to L$ as $k\to\infty$. There will be $K$ large enough such that for all $k>K$ we have that $y_k>x_{N+1}$, so $L-\varepsilon<f(x_{N+1})\le f(y_k)\le L$. 
Perhaps there are some variations of these ideas, and one can present a proof in slightly different forms. 
