Can a function $f: \mathbb{R} \to \mathbb{R}$ that has a vertical asymptote at $x = 5$ but is bounded on $(5,\infty)$ exist, why or why not? I think the answer is no because it would have to include the values $<5$, i.e. $(-\infty,5)$, but I'm not sure. Any help is appreciated.
 A: It depends on how you define an asymptote. If it suffices that either one-sided limit is infinite, then the function $$f(x)=\begin{cases} 
      \frac{1}{5-x} & x<5 \\
      0 & x\geq5
   \end{cases}
$$ satisfies the precise requirement. 
Otherwise, if one needs the two-sided limit to be infinite, then no such function exists, since by the definition of a limit, for every $N$, there will exist $\delta$ such that when $|x-5|\leq\delta$, $$|f(x)|\geq N.$$ In particular, $f$ is unbounded on $(5,\infty)$.
A: This is a very interesting question. I've tried to think through it carefully, but others can correct me if I have made a mistake (I am only a high school student). Also, I am aware my answer is very similar to User361424's, although I was already working on it when they posted the comment so I tried to add more elaboration. 
Suppose we have a piecewise function $f:\mathbb{R} \to \mathbb{R}$ defined such that $f(x) = 1/(x-5)$ for $x<5$, and as $f(x)=1$ for $x$ greater than or equal to $5$. 
Then $f$'s graph has a vertical asymptote at $x=5$ (the left-hand limit of $f$ at $x=5$ is $-\infty$, and we only need one of the two either-sided limits to be infinite in order to establish the existence of a vertical asymptote). 
But $f$ is also bounded for all $x>5$, because for all $x$ in that interval $f(x)$ remains at $1$ and doesn't go to plus or minus infinity.
Therefore, the answer to your question appears to be yes ... unless there are further restrictions you did not make clear in the original question (see the comment from User361424). 
