Hello and a Happy New Year!!

Let $f:[-1,1]\rightarrow \mathbb{R}$ be defined by $f(x)=\begin{cases}\frac{1}{k} & \text{ for } x\in \left (\frac{1}{k+1}, \frac{1}{k}\right ], \ k \in \mathbb{N}\\ 0 & \text{ for } x=x_0=0 \\ \frac{1}{k} & \text{ for } x\in \left [\frac{1}{k}, \frac{1}{k-1}\right ), \ k \in \mathbb{Z}\setminus \mathbb{N}_0, \ \text{ i.e. } (-k)\in \mathbb{N}\end{cases}$

I want to examine the existence of $f'(x_0), \ \ f_+'(x_0), \ \ f_-'(x_0), \ \ \lim_{\substack{x\rightarrow x_0+0, \\ x\in D}}f'(x), \ \ \lim_{\substack{x\rightarrow x_0-0\\ x\in D}}f'(x)$.

@TobiasKildetoft Can you give me another advide? I want to compute the limit $\lim_{n \to \infty} (\frac{n^5-10}{n^2+1})$ by only using the limit laws. I know that this sequence diverges, but I want show it once rigorously.

I tried dividing out $n^5$ so the numerator and denumerator would be two convergent sequences so I could put the limit inside the fraction. Like this

$\lim_{n \to \infty} (\frac{1-\frac{10}{n^5}}{\frac{1}{n^3}+\frac{1}{n^5}})$

But then the denumerator would converge to zero which is not allowed.

This is what the 12 times 12 companion matrix for the Möbius-function-as-roots-polynomial looks like:
$\left(
\begin{array}{cccccccccccc}
1-x & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1-x & -x & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 \\
1-x & 1 & -x & 1 & 1 & 0 & 1 & 1 & 0 & 1 & 1 & 0 \\
1-x & -x & 1 & -x & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 \\
1-x & 1 & 1 & 1 & -x & 1 & 1 & 1 & 1 & 0 & 1 & 1 \\
1-x & -x & -x & 0 & 1 & -x & 1 & 0 & 0 & 0 & 1 & 0 \\
1-x & 1 & 1 & 1 & 1 & 1 & -x & 1 & 1 & 1 & 1 & 1 \\

Q. Given
*x_1 + x_2 + x_3 + ... + x_k = N*
and *1 <= x_i <= S* for all *i in [1, k]*,
find the number of different solutions.

A. As far as I know, we can apply multinomial coefficient formula *(N-1)C(k-1)* to get the number of positive solutions to the above equation. But it will also include solutions where *x_i > S*. How do I exclude those?

Can I do the following:

Let *y_i = x_i - S*, then we can write the above equation as:
*y_1 + y_2 + y_3 + ... y_k = N-kS*

The number of positive solutions of the above equation will be *(N-kS-1)C(k-1)* and I'll subtract this from the above formula. …