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$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$

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.

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    $\begingroup$ You should explain what $[[x]]$ means. $\endgroup$ Mar 14 at 10:35
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    $\begingroup$ Incidentally, @Quasar, I was interested in going through Abbott's Analysis book earlier this year. Want to study together? $\endgroup$ Mar 14 at 22:17
  • $\begingroup$ @JeremyWeissmann, that would be super-helpful! Could we connect offline - my email address is quasar.chunawala@gmail.com. $\endgroup$
    – Quasar
    Mar 15 at 6:46
  • $\begingroup$ I've emailed you. You might need to check your spam! $\endgroup$ Mar 16 at 0:57
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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?

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