# $\delta$-response for challenging the $\lim_{x \to 10} \frac{1}{[[x]]} = \frac{1}{10}$ with $\epsilon=\frac{1}{2}$

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

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.

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

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?