I came across the following problem in Abbott's Understanding Analysis:
Consider the reasonable but erroneous claim that $$\lim_{x\to10} \frac{1}{[[x]]} = \frac{1}{10},$$ where $[[x]]$ refers to the greatest integer value of $x$.
Find the largest $\delta$ that represents a proper response to the challenge of $\epsilon = \frac{1}{2}$ and find the largest $\epsilon$ challenge for which there is no suitable $\delta$ response possible.
A student came to me with the solution to the first part as $\delta = 5$. The reasoning for this was the following:
$$|\frac{1}{[[x]]}-\frac{1}{10}| = |\frac{10-[[x]]}{10[[x]]}| = |\frac{[[x]]-10}{10[[x]]}| < \frac{|x-10|}{10}$$ and so we choose $\delta = 10\epsilon$, which means that for $\epsilon = \frac{1}{2}$, we have $\delta = 5$.
Now to my question on this:
If we consider $|\frac{1}{10}-\frac{1}{4}|=\frac{3}{20}$ and $|\frac{1}{10}-\frac{1}{16}|=\frac{3}{80},$ then would this suggest that there is a larger $\delta$ (namely 6) that is a proper response to the challenge of $\epsilon = \frac{1}{2}$?
I may have some fundamental misinterpretation here and so I welcome any help.
The proposed solution to the second claim is $\epsilon = \frac{1}{10}$. The reasoning is that, as before:
$$|\frac{1}{[[x]]}-\frac{1}{10}|<\frac{|x-10|}{10}.$$ So, picking $\epsilon = \frac{1}{10}$ will do the trick.
I don't feel right about this answer either since $\delta = 1$ will yield $$|\frac{1}{[[x]]}-\frac{1}{10}|$$ as either $\frac{1}{90}$ or $0$ since $|\frac{1}{10}-\frac{1}{10}| = 0$ and $|\frac{1}{10}-\frac{1}{9}| = \frac{1}{90}$.
Again, I might have some sort of fundamental misunderstanding here.
I appreciate any help that can be offered. Thanks in advance!