Proof of something that doesn't exist Let $\lfloor x \rfloor$ be the greatest integer function.  Show that the $\lim_{x\to 2} \frac{1}{\lfloor x \rfloor}$ does not exist.
So far I have:
Assume the limit exists.  Choose $\epsilon =1>0$.  Then for this $\epsilon$, there exists a $\delta>0$ such that if $x_1$ and $x_2$ satisfy $0<|x-0|<\delta$, then by the 2 Point Lemma the absolute value of $f(x_1)-f(x_2)<\epsilon=1$.  Since $\delta>0$, but the Archimedian property, there exists some $n \in \mathbb{N}$ such that $-\delta<-1/n<0<1/n<\delta$.  Choose $x_1=1/n$ and $x_2=-1/n$.  Thus $|\lfloor n\rfloor-(-\lfloor n \rfloor)|=|2\lfloor n \rfloor|\ge 2$.  This produces the contradiction that $2<1$, thus the limit does not exist.
Is this right, wrong, anybody indifferent?
 A: I don't see any fault with your approach, but I think you might be overworking this problem. 
In general, if you want to prove a limit does not exist, a good idea might be to prove the lateral limits are different. On this case, $\lim_{x\to 2^+} = 1/2$ and $\lim_{x\to 2^-} = 1$, as you can promptly make rigorous using $\epsilon-\delta$'s (any $\delta < 1$ will work on each side). Thus, the general limit does not exist.
A: We will write a quite (excessively?) formal $\epsilon$-$\delta$ proof. First let us note that if $x$ is a little above $2$, then $\frac{1}{\lfloor x\rfloor}=\frac{1}{2}$, while if $x$ is a little below $2$, then $\frac{1}{\lfloor x\rfloor}=1$, so it is obvious that the limit does not exist.
Now for the formal argument, which when we unwind it only says that the supposed limit $b$ cannot be simultaneously very close to $1$ and to $\frac{1}{2}$. (By the way, that means that your $\epsilon=1$ cannot work.)
Suppose to the contrary that the limit exists and is equal to $b$. Let $\epsilon=\frac{1}{10}$. Then there is a $\delta\gt 0$ such that if $|x-2|\lt \delta$, then $|f(x)-b|\lt \frac{1}{10}$, where $f(x)$ is our function.
Let $\lambda=\min(\delta/2, 1/3)$. 
Let $x_1=2-\lambda$. Then $f(x_1)=1$. Let $x_2=2+\lambda$. Then $f(x_2)=\frac{1}{2}$. 
By the Triangle Inequality, we have
$$\frac{1}{2}=|f(x_2)-f(x_1)|\le |f(x_2)-b|+|b-f(x_1)|\lt \frac{1}{10}+\frac{1}{10}=\frac{1}{5}.$$
We have reached a contradiction, since in fact it is not true that $\frac{1}{2}\lt \frac{1}{5}$. 
A: I like all the fancy proof and all, but wouldn't this be simplest if you just evaluate left-side and right-side limits? That being said, the function is discontinuous for any integer value of X. 
