Since $f$ is continuous on $(a,b)$, bounded and increasing, there's a unique continous extension of $f$ to $[a,b]$. This works because both limits $f(b) := \lim_{x\to b-}$ and $f(a) = \lim_{x\to a+}$ and are guaranteed to exist since every bounded and increasing (respectively bounded a decreasing) sequence converges. To prove this, simply observe that for a increasing and bounded sequence, all $x_m$ with $m > n$ have to lie within $[x_n,M]$ where $M=\sup_n x_n$ is the upper bound. Add to that the fact that by the very definition of $\sup$, there are $x_n$ arbitrarily close to $M$.
You can then use the fact that continuity on a compact set implies uniform continuity, and you're done. This theorem, btw, isn't hard to prove either (and the proof shows how powerful the compactness property can be). The proof goes like this:
First, recall the if $f$ is continuous then the preimage of an open set, and in particular of an open interval, is open. Thus, for $x \in [a,b]$ all the sets $$
C_x := f^{-1}\left(\left(f(x)-\frac{\epsilon}{2},f(x)+\frac{\epsilon}{2}\right)\right) $$
are open. The crucial property of these $C_x$ is that for all $y \in C_x$ you have $|f(y)-f(x)| < \frac{\epsilon}{2}$ and thus $$
|f(u) - f(v)| = |(f(u) - f(x)) - (f(v)-f(x))|
\leq \underbrace{|f(u)-f(x)|}_{<\frac{\epsilon}{2}}
+ \underbrace{|f(v)-f(x)|}_{<\frac{\epsilon}{2}} < \epsilon
\text{ for all } u,v \in C_x
$$
Now recall that an open set contains an open interval around each of its points. Each $B_x$ thus contains an open interval around $x$, and you may wlog assume that its symmetric around $x$ (just make it smaller if it isn't). Thus, there are $$
\delta_x > 0 \textrm{ such that }
B_x := (x-\frac{\delta_x}{2},x+\frac{\delta_x}{2})
\subset (x-\delta_x,x+\delta_x)
\subset C_x
$$
Note how we made $B_x$ artifically smaller than seems necessary, that will simplify the last stage of the proof. Since $B_x$ contains $x$, the $B_x$ form an open cover of $[a,b]$, i.e. $$
\bigcup_{x\in[a,b]} B_x \supset [a,b] \text{.}
$$
Now we invoke compactness. Behold! Since $[a,b]$ is compact, every covering with open sets contains a finite covering. We can thus pick finitely many $x_i \in [a,b]$ such that we still have $$
\bigcup_{1\leq i \leq n} B_{x_i} \supset [a,b] \text{.}
$$
We're nearly there, all that remains are a few applications of the triangle inequality. Since we're only dealing with finitly many $x_i$ now, we can find the minimum of all their $\delta_{x_i}$. Like in the definition of the $B_x$, we leave ourselves a bit of space to maneuver later, and actually set $$
\delta := \min_{1\leq i \leq n} \frac{\delta_{x_i}}{2} \text{.}
$$
Now pick arbitrary $u,v \in [a,b]$ with $|u-v| < \delta$.
Since our $B_{x_1},\ldots,B_{x_n}$ form a cover of $[a,b]$, there's an $i \in {1,\ldots,n}$ with $u \in B_{x_i}$, and thus $|u-x_i| < \frac{\delta_{x_i}}{2}$. Having been conservative in the definition of $B_x$ and $\delta$ pays off, because we get $$
|v-x_i| = |v-((x_i-u)+u)| = |(v-u)-(x_i-u)|
< \underbrace{|u-v|}_{<\delta\leq\frac{\delta_{x_i}}{2}}
+ \underbrace{|x_i-u|}_{<\frac{\delta_{x_i}}{2}}
< \delta_{x_i} \text{.}
$$
This doesn't imply $y \in B_{x_i}$ (the distance would have to be $\frac{\delta_{x_i}}{2}$ for that), but it does imply $y \in C_{x_i}$!. We thus have $x \in B_{x_i} \subset C_{x_i}$ and $y \in C_{x_i}$, and by definition of $C_x$ (see the remark about the crucial property of $C_x$ above) thus $$
|f(x)-f(y)| < \epsilon \text{.}
$$
Induction isn't really necessary here. It is enough to just work with arbitrary $n$ to show the argument works for whatever $n$ you choose.
Indeed, fix $\varepsilon > 0$. We want $|x^{n+1} - x_0^{n+1}| < \varepsilon$. Notice that
$$|x^{n+1} - x_0^{n+1}| = |x - x_0||x^n + x^{n-1}x_0 + \dots + x x_0^{n-1} + x_0^n|$$
When $x$ is in an interval around $x_0$, say $x \in [x_0 - 1, x_0+1]$, the sum on the RHS has a bound, call this bound $M$.
Hence, taking $\delta = \min(\dfrac{\varepsilon}{M}, 1)$ we get
$$|x^{n+1} - x_0^{n+1}| = |x - x_0||x^n + x^{n-1}x_0 + \dots + x x_0^{n-1} + x_0^n| < \delta M = \varepsilon$$
so we are done.
Best Answer
The answer was essentially given. Here is an explicit way to connect the ideas.
Take $\epsilon=1$ in the definition of limit. So,
due to $\lim_{x\to-\infty}=L_1$, there exists $N_1$ such that $f(x)\in(L_1-1,L_1+1)$ for all $x<N_1$.
due to $\lim_{x\to\infty}=L_2$, there exists $N_2>N_1$ such that $f(x)\in(L_2-1,L_2+1)$ for all $x>N_2$.
Finally, due to continuity, $f$ is bounded on the compact interval $[N_1,N_2]$. Hence, there exists $C$ such that $|f(x)|\leq C$ for all $x\in[N_1,N_2]$.
So, you can take $M=\max\{|L_1-1|,|L_1+1|, |L_2-1|, |L_2+1|,C\}$ to conclude that $|f(x)|<M$ for all $x\in\mathbb{R}={\rm Dom}(f)$.