Challenging $\lim_{x \rightarrow 10} \frac{1}{\lfloor x \rfloor} = \frac{1}{10}$ for $\epsilon=\frac{1}{2}$. Consider the (incorrect) claim that $$\lim_{x \rightarrow 10} \frac{1}{\lfloor x \rfloor} = \frac{1}{10}.$$
How might I find the largest $\delta$ such that I can challenge $\epsilon = 1/2$?
Clearly I need to use the $\epsilon$-$\delta$ definition of a limit, but my problem is that I don't know how to find the largest delta.
Here is what I know:


*

*$|f(x) - (1/10)| <1/2$

*$0<|x-10|< \delta$


Thinking aloud, so your limit $L=\dfrac1{10}$ and your function $f(x)=\dfrac1{\lfloor x\rfloor}$.  
$\forall\varepsilon\exists\delta$ etc. So, for $\varepsilon=\dfrac12$, $\exists\delta$ such that if $|x-10|<\delta$ then $\bigl|\dfrac1{\lfloor x\rfloor}-\dfrac1{10}\bigr|<\dfrac12$.  
That is $\dfrac{-1}2 + \dfrac1{10}< \dfrac1{\lfloor x\rfloor} < \dfrac12 + \dfrac1{10}$, or equivalently $\dfrac{-4}5< \dfrac1{\lfloor x\rfloor} < \dfrac35$ .   
If $x>0$ (or rather when $x\ge1$) then we need $\lfloor x\rfloor>\dfrac53$, so $\lfloor x\rfloor\ge2$.
When $x<0$ we need $\dfrac45> \dfrac{-1}{\lfloor x\rfloor}$, so $\dfrac54< -\lfloor x\rfloor$, in turn 
$2\le -\lfloor x\rfloor$, thus $-2\ge \lfloor x\rfloor$, in turn $x<-1$. 
How do we ensure that either $\lfloor x\rfloor\ge2$ or $x<-1$, using the single condition $|x-10|<\delta$?
The latter reads as $-\delta+10<x<\delta+10$.
It looks to me we could take $\delta$ as big as $8$, and the condition 
$-\delta+10<x<\delta+10$ becomes $2<x<18$, which implies $\lfloor x\rfloor\ge2$, which implies $\bigl|\dfrac1{\lfloor x\rfloor}-\dfrac1{10}\bigr|<\dfrac12$.
But $\delta$ cannot be bigger than $8$ as then $x<2$ will be allowed, 
but that would entail $\lfloor x\rfloor=1$ (for at least some $x$) but then 
$\bigl|\dfrac1{\lfloor x\rfloor}-\dfrac1{10}\bigr| = 1- \dfrac1{10} = \dfrac9{10}\ge\dfrac12$, which is a problem, not permitted. 
So, I would suggest $\delta=8$, post this, and keep reading to see if I have mistakes. 
Edit. On second reading I believe $\delta =8$ is naturally related to your question, but I do not understand your question. What would mean for $\delta$ to challenge $\varepsilon$? If I interpret this the only way I am capable of, it seems that my $\delta = 8$ would be the smallest that does not challenge the $\varepsilon=\dfrac12$, and then there is no largest $\delta$ to challenge the $\varepsilon$, for all $\delta>8$ work. 
Edit$^2$. Perhaps I wanted to say that $\delta = 8$ would be the  smallest largest that does not challenge the $\varepsilon=\dfrac12$. 
