Plotting $f(x) = x\lfloor 1/x \rfloor$

Question

I want to plot $f(x) = x\lfloor 1/x \rfloor$ near the point zero for finding its limit but I can't choose proper intervals and plot it .

Answer 1

This is what I obtain from GeoGebra:

No matter how much I zoomed in, I eventually ended up with a similar looking pattern so I suppose there is no 'appropriate' interval to plot this function.

Using Wolfram|Alpha, I get the series representation $$x\left\lfloor{\frac 1x}\right\rfloor=1 - \frac x2 + \frac x\pi \sum_{k=1}^\infty \frac{\sin\left(\frac{2 k \pi}{x}\right)}{k}$$ which confirms the suspicion that $$\lim_{x\to0}x\left\lfloor{\frac 1x}\right\rfloor=1$$ but I have no idea how the series is derived.

Did some research just now and it looks like the series can be quite easily obtained.

Answer 2

On an unrelated note, it's easier to calculate the limit than you think. Since $k - 1\le\lfloor k \rfloor \le k $, we know that $$\lim_{x \to 0}x\left(\frac{1}x - 1\right)\le \lim_{x \to 0}x\left\lfloor {1 \over x}\right\rfloor \le \lim_{x \to 0}x \left( \frac{1}x\right)$$Now you see the squeeze theorem in action. It's not exactly a good idea to make an exact plot of this function by hand, since you've got all integers covered up as you approach zero, but you may tentatively understand that for large numbers, $\lfloor x\rfloor$ may be treated as $x$, the latter being continuous instead of discrete.

Answer 3

When $\frac{1}{n+1} < x \leq \frac{1}{n}$ we have $n \leq \frac{1}{x} < n+1$ and thus $\lfloor \frac{1}{x} \rfloor = n$ and $x \lfloor \frac{1}{x} \rfloor = nx.$ The range on the interval $\frac{1}{n+1} < x \leq \frac{1}{n}$ will then be $(\frac{n}{n+1}, \frac{n}{n}] = (1-\frac{1}{n+1}, 1].$ From this it's obvious that the limit is $1.$

Answer 4

Check the graph at wolfram

The the limit is 1 as you can see in the graph.

If you want to explain this in a more rigorous way, let's apply limit definition.

Given any $\epsilon >0$ you can find an $x_0$ such as $|f(x) - 1| < \epsilon$ for all $x < x_0$. Let's just work with the limit from the right, for simplicity:

$$ |x_0 [\frac{1}{x_0}] - 1 |< \epsilon$$

We also know that $[x] = x - \{x\}$, where $\{\}$ means the fractional part. So we can rewrite the former expression as:

$$|x_0 (\frac{1}{x_0} - \{\frac{1}{x_0}\}) -1| < \epsilon$$

Since we are operating from the right, we just need to swap the sign to get rid of absolute value:

$$ 1- x_0 (\frac{1}{x_0} - \{\frac{1}{x_0}\}) < \epsilon$$

$$ x_0\{\frac{1}{x_0}\} < \epsilon$$

The worst case scenario is when fractional part of $x_0$ is as close as posible to $1$, taking $x_0 < \epsilon$ assures the limit definition is always followed.

So we can conclude

$$\lim_{x-> 0^{+}}{x[\frac{1}{x}]} = 1$$

Similar you can proceed to calculate the limit from the left, just take care with the absolute value function.

Answer 5

I switched to $1/x$ for large positive $x,$ picture gives better idea this way. I would have preferred using a different letter, say $t = 1/x,$ but I do not yet know how to do that. So, think of the graphs as indicating that the limit as $t \rightarrow +\infty \; \; \; \lfloor t \rfloor /t = 1,$ and $t \rightarrow -\infty \; \; \; \lfloor t \rfloor /t = 1.$ That can then be applied to $x = 1/t$ at $0^+$ and $0^-$

Apparently the intended question is to apply the floor function again. In the graphs below, one can see easily what happens in the positive case and the negative case, when $t$ is an integer and when $t$ is not an integer. Just look at the graphs, think of what the floor function does if applied that one more time. Note: I did try writing an extra floor function in the graph, the computer did not do it correctly.

Next for negative $x$