I am working through Understanding Analysis 2nd Ed. and I am having some difficulty working with functional limits involving floor functions.

Exercise 4.2.4: Consider the reasonable but erroneous claim that $\lim_{x \to 10} \frac{1}{\lfloor x \rfloor} = \frac{1}{10}$

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

Note: $x-1 < \lfloor x \rfloor \leq x$

Rough work:

$$0 < \lvert x - 10 \rvert < \delta \rightarrow \lvert \frac{1}{ \lfloor x \rfloor} - \frac{1}{10} \rvert < \epsilon$$

$$\frac{1}{10} - \epsilon < \frac{1}{ \lfloor x \rfloor} < \epsilon + \frac{1}{10}$$

$$-\frac{2}{5} < \frac{1}{ \lfloor x \rfloor} < \frac{3}{5}$$

$$ \lfloor x \rfloor < -\frac{5}{2} \quad and \quad \lfloor x \rfloor > \frac{5}{3}$$

$$ x < -3 \quad and \quad x > 2$$

I'm looking for some way to formulate $0 < \lvert x - 10 \rvert < \delta$ again in order to determine a $\delta$. Intuitively looking at this I don't seen how a delta-ball exists because of the disjoint-ness of the inequality. This makes me think that I either 1) incorrectly applied the inversion of $\frac{1}{ \lfloor x \rfloor}$ to the inequality, or 2) this is not a good approach to solving the problem.

I've been able to solve some floor function limits intuitively (just looking at the behavior in general), but I'd like to be more rigorous in my explanations. Any guidance here would be greatly appreciated!

  • $\begingroup$ There is no "smooth" approach to $\frac{1}{10}$ with this function, it presents with jumps between integer values of $x$ (step-wise I think). However, I am being asked to find the largest $\delta$ for $\epsilon = \frac{1}{2}$. This combined with behavior of the function (how to work with it mathematically) is confusing for me with my current knowledge of math. $\endgroup$ Jul 13 at 13:49
  • $\begingroup$ I misunderstood the question. I've deleted my comments. I suggest that you delete your responses, to prevent confusion. $\endgroup$
    – saulspatz
    Jul 13 at 13:57
  • $\begingroup$ I am liking this question less and less. I think if I have $x_{n} = 10 - \frac{1}{n}$ and $y_{n} = 10 + \frac{1}{n}$ then $\lim_{x \to 10} f(x_{n}) = 9$ and $\lim_{x \to 10} f(y_{n}) = 10$, which clearly shows discontinuity (intuitively this is clear as well). I don't really understand how to do what this problem wants me to do $\endgroup$ Jul 13 at 13:58
  • $\begingroup$ I don't understand the point of finding largest $\delta$. Any value of $\delta$ which works is fine. Any emphasis on finding optimal $\delta$ is so so against the understanding of limit definition. $\endgroup$ Jul 15 at 15:33

You're already halfway there to find your $\delta$. Notice you've already established that $$\color{purple}{2 <x} \implies \Big\lvert \frac{1}{\lfloor x \rfloor} - \frac{1}{10} \Big \rvert < \frac{1}{2}$$ And $$ \lvert x - 10 \rvert < \delta \iff \color{purple}{10 - \delta < x} < 10 + \delta $$ So taking $\delta = 10 - 2 = 8$ will guarantee the desired property.

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  • $\begingroup$ Okay, thanks! So this problem is asking us to be naive to the requirement $x < -\frac{5}{2}$? $\endgroup$ Jul 13 at 14:14
  • 1
    $\begingroup$ Good question. You don't choose to ignore it, but you can't get a solution if you try to use it. If we take $10 + \delta = -3 \implies \delta = -13$, but by definition $\delta >0$ so this case is impossible. $\endgroup$
    – Robert Lee
    Jul 13 at 14:19
  • $\begingroup$ Your inequality $\lfloor x \rfloor < -5/2$ is incorrect. It looks to me like when you derived it, you forgot about the fact that if you multiply both sides of an inequality by a negative number, you have to reverse the direction of the inequality. $\endgroup$ Jul 13 at 17:21
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    $\begingroup$ @DanVelleman. If $\lfloor x\rfloor <0$ then $-\frac{2}{5} <\frac{1}{\lfloor x\rfloor} \iff -\frac{5}{2}>\lfloor x \rfloor$ (as explained here), so $$-\frac{2}{5} < \frac{1}{ \lfloor x \rfloor} \iff \lfloor x \rfloor \in \left(-\infty, - \frac{5}{2}\right) \cup (0, \infty)$$ $\endgroup$
    – Robert Lee
    Jul 13 at 17:50
  • $\begingroup$ @Robert Lee: Thanks for the explanation. It's hard to know what OP was thinking. If this was OP's reasoning, then "$\lfloor x \rfloor < -5/2$ and $\lfloor x \rfloor > 5/3$" in the original post should have been "$\lfloor x \rfloor < -5/2$ or $\lfloor x \rfloor > 5/3$." $\endgroup$ Jul 14 at 0:09

You calculations are correct up to the line $$-\frac{2}{5} < \frac{1}{ \lfloor x \rfloor} < \frac{3}{5}\tag1$$

When $x\geq10$, we have $\lfloor x\rfloor\geq10$ so that$$0\leq\frac1{\lfloor x\rfloor}\leq\frac1{10}$$ and $(1)$ is certainly satisfied, so we must consider $x<10$. Since $0$ is not in the domain of our function, we must restrict our attention to positive $x$, and since $\lfloor x\rfloor$ is always an integer, we need to find the largest positive integer $n$ such that $\frac1n-\frac1{10}<\frac12$. Clearly, the largest such $n$ is $2$, and $\delta = 10-2=8$.


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