I'm having trouble with the following problem.
Given the standard greatest integer function $\lfloor x \rfloor = int(x)$ where $ \lfloor x \rfloor $ returns the greatest integer less than or equal to x. I must find the largest $\delta$ such that when $ | x - 10 | < \delta$ it follows that $ | \frac{1}{\lfloor x \rfloor} - \frac{1}{10} | < 1/2 $.
Also I must find the largest $ \epsilon $ such that no matter the $ \delta $ chosen $|\frac{1}{\lfloor x \rfloor} - \frac{1}{10}| >= 1/2$.
I tried tackling the problem by considering an algebraic manipulation of $|x - 10|$ into $ | \frac{1}{\lfloor x \rfloor} - \frac{1}{10} | $ to find a suitable $ \delta $ but I cant seem to figure out how to do this. Everything looks like its been raised to -1 so I suspect $ \delta $ = 2 but I do not know how to rigorously show that. Any help?
First think "what are the possible integers $n$ so that
$$\left| \frac{1}{n} - \frac{1}{10}\right| < 1/2?$$
Well, on one end of the spectrum we know that $n = 2$ works but $n=1$ does not. Furthermore, since, if $n$ is positive, we have
$$\left|\frac{1}{n} - \frac{1}{10}\right| < \frac{1}{10} < 1/2$$
we know all positive $n \geq 2$ works.
Hence, we want $\lfloor x \rfloor \geq 2$. This is precisely when $x \geq 2$, so the largest interval around $10$ we can fit inside this range is one of radius $8$, namely the interval $[2, 18]$.
Also, your second sentence doesn't make sense: you say the largest $\varepsilon$, but $\varepsilon$ isn't a part of that equation. If you mean $\varepsilon$ instead of $1/2$, then we can solve it like this: if we take $\delta$ small enough, we can make it so that either $\lfloor x \rfloor = 10$ or $\lfloor x \rfloor = 9$ in that interval. Furthermore, this is the best we can do, no matter how small of a $\delta$ we take. The maximum difference in this case is $|1/10 - 1/9| = 1/90$.