Challenging $\lim_{x \rightarrow 10} \frac{1}{\lfloor x \rfloor} = \frac{1}{10}$ for $\epsilon=\frac{1}{2}$.

Question

Consider the (incorrect) claim that $$\lim_{x \rightarrow 10} \frac{1}{\lfloor x \rfloor} = \frac{1}{10}.$$

How might I find the largest $\delta$ such that I can challenge $\epsilon = 1/2$?

Clearly I need to use the $\epsilon$-$\delta$ definition of a limit, but my problem is that I don't know how to find the largest delta.

Here is what I know:

• $|f(x) - (1/10)| <1/2$
• $0<|x-10|< \delta$

Answer 1

You cannot refute existence of limit based on error margin of $ε = \frac{1}{2}$, because the function is indeed close to $\frac{1}{10}$ within that error margin in a suitably small interval around $10$. If you want to disprove the existence of the limit, you need a smaller error margin. Just explicitly calculate what $\frac{1}{\lfloor x \rfloor}$ is when $x \in (9,10)$ and when $x \in (10,11)$ and you will know what error margin you need.

Answer 2

Thinking aloud, so your limit $L=\dfrac1{10}$ and your function $f(x)=\dfrac1{\lfloor x\rfloor}$.

$\forall\varepsilon\exists\delta$ etc. So, for $\varepsilon=\dfrac12$, $\exists\delta$ such that if $|x-10|<\delta$ then $\bigl|\dfrac1{\lfloor x\rfloor}-\dfrac1{10}\bigr|<\dfrac12$.

That is $\dfrac{-1}2 + \dfrac1{10}< \dfrac1{\lfloor x\rfloor} < \dfrac12 + \dfrac1{10}$, or equivalently $\dfrac{-4}5< \dfrac1{\lfloor x\rfloor} < \dfrac35$ .

If $x>0$ (or rather when $x\ge1$) then we need $\lfloor x\rfloor>\dfrac53$, so $\lfloor x\rfloor\ge2$.
When $x<0$ we need $\dfrac45> \dfrac{-1}{\lfloor x\rfloor}$, so $\dfrac54< -\lfloor x\rfloor$, in turn $2\le -\lfloor x\rfloor$, thus $-2\ge \lfloor x\rfloor$, in turn $x<-1$.

How do we ensure that either $\lfloor x\rfloor\ge2$ or $x<-1$, using the single condition $|x-10|<\delta$?
The latter reads as $-\delta+10<x<\delta+10$.
It looks to me we could take $\delta$ as big as $8$, and the condition $-\delta+10<x<\delta+10$ becomes $2<x<18$, which implies $\lfloor x\rfloor\ge2$, which implies $\bigl|\dfrac1{\lfloor x\rfloor}-\dfrac1{10}\bigr|<\dfrac12$.
But $\delta$ cannot be bigger than $8$ as then $x<2$ will be allowed, but that would entail $\lfloor x\rfloor=1$ (for at least some $x$) but then $\bigl|\dfrac1{\lfloor x\rfloor}-\dfrac1{10}\bigr| = 1- \dfrac1{10} = \dfrac9{10}\ge\dfrac12$, which is a problem, not permitted.

So, I would suggest $\delta=8$, post this, and keep reading to see if I have mistakes.

Edit. On second reading I believe $\delta =8$ is naturally related to your question, but I do not understand your question. What would mean for $\delta$ to challenge $\varepsilon$? If I interpret this the only way I am capable of, it seems that my $\delta = 8$ would be the smallest that does not challenge the $\varepsilon=\dfrac12$, and then there is no largest $\delta$ to challenge the $\varepsilon$, for all $\delta>8$ work.

Edit$^2$. Perhaps I wanted to say that $\delta = 8$ would be the smallest largest that does not challenge the $\varepsilon=\dfrac12$.

Answer 3

Since your question is moderately unclear at the moment, I thought I would give you some thoughts that may be helpful in forging ahead.

How might I find the largest $\delta$ such that I can challenge $\epsilon = 1/2$?

I think this is the crux of the problem. What do you mean by "challenge" exactly? This is the part that is currently not clear about your question. First of all, what does $|x-a|$ represent? It represents the distance from $x$ to $a$. Hence, the meaning of $0<|x-a|<\delta$ should be clearer now; $\delta$ basically represents "how many units, in either direction, $x$ is permitted to stray from $a$." In your case, $a=1/10$. Now consider the following.

We must contend with $$ f(x)=\left|\frac{1}{\lfloor x\rfloor}-\frac{1}{10}\right|<\frac{1}{2}.\tag{1} $$ For what $x$-values is $(1)$ true? A graph may help:

Above, the straight line is the constant value $1/2$ for all $x$-values, and the scattered lines form a plot for your function $f(x)$ in $(1)$. Can you tell from the graph (or algebraically) for which $x$-values $(1)$ holds?

We can see (or reason) that $(1)$ holds true when $x\in(-\infty,-2)\cup[2,\infty)$ but fails if $x\in[-2,2)$. With this in mind, what is the maximum $\delta$-value for which $(1)$ is not true or "challenged"? The maximum value for $\delta$ will be achieved at that point which is farthest from $x=10$. This occurs at $x=-2$, a distance of $12$ units from $x=10$.

Hence, the largest $\delta$ for which you can challenge $\epsilon=1/2$ is $\delta=12$. That is the only reasonable way to answer this question in my opinion.