How much is $\lceil\frac{1}{\infty}\rceil$? How much is  $\lceil\frac{1}{\infty}\rceil$  ?
On one hand, $\frac{1}{\infty}=0$, so its ceiling is also $0$.
On the other hand, for all $x\geq 1$, $\lceil\frac{1}{x}\rceil = 1$, so, when $x$ goes to infinity, the function should remain with the same value...
 A: It's always dangerous to write $\infty$ in calculations. You have to be sure what you mean with writing $\infty$. In this case, you have two possibilities:
$$\lim_{n\to\infty}\left\lceil \frac{1}{n} \right\rceil=1$$ or
$$\left\lceil \lim_{n\to\infty} \frac{1}{n} \right\rceil=0.$$
You can not switch a function and a limit without further explanation. Compare with the important problem of the analysis where they try to switch a limit and an integral (the reason why the Lebesgue-integral is invented).
A: Great question!  The main problem anyone's going to have answering it is that the fraction $\frac1\infty$ isn't defined: $\infty$ isn't part of the ordered field of real numbers, and you can't just divide $1$ by $\infty$.  Moreover, the function $\lceil.\rceil$ is only usually defined on the real number line, so even in strange fields like the hyperreal numbers (I think), where things like $\frac1\infty$ do make sense, it's a lot harder to make sense of $\lceil\frac1\infty\rceil$.  
Here's one way you might try to make sense of it.  You say that $\frac1\infty=0$.  In mathematics, we don't say that, but we do say that $\lim_{x\to\infty}\frac1x=0$.  '$\lim_{x\to\infty}$' means that we consider the behaviour of $\frac1x$ as $x$ becomes arbitrarily large.  As we make $x$ larger, $\frac1x$ gets closer and closer to $0$: in fact, we can make it as close to $0$ as we like by choosing $x$ large enough.  So $\lim_{x\to\infty}\frac1x=0$.  Therefore, $\lceil\lim_{x\to\infty}\frac1x\rceil=0$.  
But we could also interpret $\lceil\frac1\infty\rceil$ as the mathematical expression $\lim_{x\to\infty}\lceil\frac1x\rceil$.  Now we are considering the behaviour of $\lceil\frac1x\rceil$ as $x$ becomes arbitrarily large.  This time, the behaviour is much simpler: as long as $x\ge1$, $\lceil\frac1x\rceil=1$.  Therefore, $\lim_{x\to\infty}\lceil\frac1x\rceil=1$.  
Because we can't swap the order of taking limits and taking the ceiling function, we say that the ceiling function $\lceil y\rceil$ is discontinuous at the point $y=0$.  Discontinuity means that there is a 'jump' in the graph of the function: you can't draw it without taking your pen off the page: 

Notice the 'jump' at the point $0$ in this graph of the ceiling function.  
The opposite of 'discontinuous' is continuous.  A function $f$ is continuous at a point $a$ if $f(a)=\lim_{x\to a}f(x)$.  Functions which are continuous everywhere include all polynomials, and lots of other lovely functions like $\sin$, $\cos$ and the Bessel functions, but the ceiling function is discontinuous at $0$, which is why we can't give meaning to $\lceil\frac1\infty\rceil$.  
A: To elaborate on the comment above by @Jonas Meyer, there are number systems extending $\mathbb{R}$ which contain infinite numbers.  If the symbol "$\infty$" is interpreted as referring to such a positive number, then $\frac1\infty$ is a positive infinitesimal. In some of these number systems such as the hyperreals, there is a principle that allows one to extend functions to the larger number system.  In particular, the floor and ceiling functions extend in this way, and one neccesarily has that $\lceil\frac1\infty\rceil$ is indeed 1. 
A: This is a great example of why we can't pass limits inside of non-continuous functions.  Notice the following:
$$1=\lim_{n\to\infty}1=\lim_{n\to\infty}\left\lceil\frac{1}{n}\right\rceil\\0=\lceil 0\rceil=\left\lceil\lim_{n\to\infty}\frac{1}{n}\right\rceil$$
A: No, $\frac1\infty$ is not $0$: it is undefined. So, therefore, is $\left\lceil\frac1\infty\right\rceil$.
Your argument that it ought to be $1$ is also incorrect: it’s based on an unconscious assumption that the ceiling function is continuous. The same argument would say that since $1=\lim_{x\to 0^+}(x+1)$, and since $\lceil x+1\rceil\ge 2$ for all $x>0$, with $\lceil x+1\rceil=2$ for all small $x>0$, therefore $\lceil x+1\rceil$ ought to be $2$. But of course it isn’t: it’s $1$. The ceiling function isn’t continuous from the right at integers.
A: $\frac{1}{\infty}=lim_{x\to\infty}\frac{1}{x}=lim_{x\to 0^+}x=0$ and ceil function is right discontinuous at $\mathbb{Z}$ especially at zero,now We have $\lceil\frac{1}{\infty}\rceil=0=\lceil\lim_{x\to 0^+}x\rceil\neq lim_{x\to 0^+}\lceil x\rceil=1$.
