Why does this limit $\lim_{x\to b} f(x)$ exist? Let $f:(a,b) \to \mathbb R$ be a continuous injective function. 
I want to continuously extend it to $[a,b]$. To do so I need $\lim_{x\to a}f(x)$ and $\lim_{x\to b}f(x)$ to exist. But it's not clear to me that they exist.

Why does $\lim_{x\to b}f(x)$ exist if $f$ is injective?

 A: These limits might not exist. Let $a=-\pi/2$, $b=\pi/2$, and $f(x)=\tan(x)$. You need a stronger form of continuity for these limits to exist: uniform continuity.
A: You have two ways here.
First: Assume $f$ is bounded
By doing that, you have nice, real $\sup$ and $\inf$ of the function.
It is not so difficult to prove, via the intermediate value theorem, that a continuous injective function on an interval is either strictly decreasing or strictly decreasing. WLOG, we can assume it is increasing, and it is then easy to prove that $\lim\limits_{x \to b}f(x)=\sup\limits_{x \in (a,b) } f$, and $\lim\limits_{x \to a} f(x)=\inf\limits_{x \in (a,b) } f$.

Second: Enlarge your codomain
If you consider the extension of $f$ to the extended number line, that is, $f:(a,b) \rightarrow [-\infty,\infty]$, then $\sup$ and $\inf$ will exist, regardless of $f$ being bounded or not. And you can extend $f$ to a continuous function on $[a,b]$ either way, noting that you can end up having $f(a)$ or $f(b)$ being $\infty$ or $-\infty$.

Now, answering your question directly: The limits may not exist if $f$ is not bounded. If it is, they do, by what I said in "First". If $f$ is not bounded, they may not exist, but you can "fix" the problem using what I said in "Second".
A: $f$ must be uniformly continuous so that it can be extended continuously to $[a,b] = (a,b)^-$. (Closure)
The limits will exist but might be $\pm \infty$.
