<s>[INST] $\delta$-response for challenging the $\lim_{x \to 10} \frac{1}{[[x]]} = \frac{1}{10}$ with $\epsilon=\frac{1}{2}$ $\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$
I am self-studying Real Analysis from Stephen Abbott's Understanding Analysis. I'd like to ask if my conclusions pertaining to exercise (a), (b) of the below problem are correct.
Notation. $f(x)= [[x]]$ is the box function, the greatest integer less than or equal to $x$, for all $x \in \mathbf{R}$.

Exercise 4.2.4. Consider the reasonable but erroneous claim that

\begin{align*}
\lim_{x \to 10} \frac{1}{[[x]]} = \frac{1}{10}
\end{align*}

(a) Find the largest $\delta$ that represents a proper response to the challenge of $\epsilon = 1/2$.


(b) Find the largest $\delta$ that represents a proper response to $\epsilon = 1/50$.


(c) Find the largest $\epsilon$ challenge for which there is no suitable $\delta$ response possible.

Proof.
(a) We require
\begin{align*}
\frac{1}{10} - \frac{1}{2} &< \frac{1}{[[x]]} &< \frac{1}{10} + \frac{1}{2} \\
\frac{-4}{10} &< \frac{1}{[[x]]} &< \frac{6}{10}
\end{align*}
So, $[[x]] < \frac{-10}{4} < -2$ and $[[x]]>\frac{10}{6} > 1$. In other words, $x-10 < -12$ and $x-10>-8$. The absolute distance must satisfy the inequality $\absval{x - 10} < 8$. Thus, the largest $\delta-$response to the challenge $\epsilon=1/2$ appears to be $\delta = 8$.
(b) We require
\begin{align*}
\frac{1}{10} - \frac{1}{50} &< \frac{1}{[[x]]} &< \frac{1}{10} + \frac{1}{50} \\
\frac{4}{50} &< \frac{1}{[[x]]} &< \frac{6}{50}
\end{align*}
So, $[[x]] < \frac{50}{4} < 13$ and $[[x]]>\frac{50}{6} > 8$. In other words, $x < 13$ or $x >9$. The absolute distance must satisfy the inequality $\absval{x - 10} < 1$. Thus, the largest $\delta-$response to the challenge $\epsilon=1/50$ appears to be $\delta = 1$.
(c) We would like to have the distance
\begin{align*}
\absval{\frac{1}{[[x]]} - \frac{1}{10}} > \epsilon
\end{align*}
no matter what the open interval $(10-\delta,10+\delta)$ in which $x$ lies. Rearranging, I get:
\begin{align*}
\epsilon < \frac{\absval{[[x]] - 10}}{10 \absval{[[x]]}}
\end{align*}
I am not sure how to proceed from here. I know that, $\absval{[[x]] - 10}$. That yields,
\begin{align*}
\epsilon < \frac{\delta}{10 \absval{[[x]]}}
\end{align*}
I can further write, $[[x]] > \lceil{10 - \delta}\rceil$. But, this $\epsilon$ is dependent on the $\delta$-interval I choose.
 [/INST] For part (c), we’re looking for a distance $\epsilon$ such that, when we look around x = 10, the function values aren’t all closer than $\epsilon$ to 1/10.
I feel like a more intuitive approach might be effective here. What value does the function take on to the left of x = 10? How far is that value from 1/10? That difference should be your $\epsilon$, no?
</s>