$x_n=f\left(\frac{1}{n^2}\right)+f\left(\frac{2}{n^2}\right)+\cdots+f\left(\frac{n}{n^2}\right)$, $\lim_{n\to\infty}x_n=\dfrac{f'(0)}{2}$ Let $f(x)$ be a function and $f(0)=0$, $f'(0)$ exists. Let
$$x_n=f\left(\frac{1}{n^2}\right)+f\left(\frac{2}{n^2}\right)+\cdots+f\left(\frac{n}{n^2}\right).$$
Prove that $\lim_{n\to\infty}x_n=\dfrac{f'(0)}{2}$.
The solution is given by: Since $f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}$, we have that
$$f(x)=f(0)+f'(0)x+o(x)=f'(0)x+o(x).$$
When $n\to+\infty$, we have that
$$f\left(\frac{1}{n^2}\right)=\frac{f'(0)}{n^2}+o\left(\frac{1}{n^2}\right)=
\frac{f'(0)}{n^2}+o\left(\frac{1}{n}\right),\quad
f\left(\frac{2}{n^2}\right)=\frac{2f'(0)}{n^2}+o\left(\frac{2}{n^2}\right)=\frac{2f'(0)}{n^2}+o\left(\frac{1}{n}\right)$$
$$f\left(\frac{k}{n^2}\right)=\frac{kf'(0)}{n^2}+o\left(\frac{k}{n^2}\right)=
\frac{kf'(0)}{n^2}+o\left(\frac{1}{n}\right),\dots$$
Hence
$$x_n=f'(0)\cdot\frac{1+2+\cdots+n}{n^2}+n\cdot o\left(\frac{1}{n}\right)=f'(0)\cdot\frac{n+1}{2n}+o(1),$$
then
$$\lim_{n\to\infty}x_n=\frac{f'(0)}{2}.$$
I have a question: in the above steps, $o\left(\frac{1}{n}\right)$ are different, can we do this $n\cdot o\left(\frac{1}{n}\right)=o(1)$.
A more accurate question is: suppose there is a two-dimensional sequence $\{x_{nm}\}_{n,m\in\mathbb{N}}$ and for a fixed $m$, there will always be $\lim_{n\to\infty}x_{nm}=0$, is it right that
$$\lim_{n\to\infty}\frac{x_{n1}+x_{n2}+\cdots+x_{nn}}{n}=0 ?$$
The original intention of this question is that we used the conclusion $n\cdot o(1/n)=o(1)$ in the answers to the above questions. Since these $o(1/n)$ are different, set to
$$\alpha_{n1},\alpha_{n2},\dots,\alpha_{nn},\quad \alpha_{nk}=o(1/n),\quad 1\leq k\leq n.$$
which is
$$\lim_{n\to\infty}\dfrac{\alpha_{nk}}{\dfrac{1}{n}}=\lim_{n\to\infty}n\alpha_{nk}=0,\quad 1\leq k\leq n.$$
Is the limit of the sum of $\alpha_{n1},\alpha_{n2},\dots,\alpha_{nn}$ is also $0$? That is whether there is
$$\lim_{n\to\infty}(\alpha_{n1}+\alpha_{n2}+\cdots+\alpha_{nn})=
\lim_{n\to\infty}\frac{n\alpha_{n1}+n\alpha_{n2}+\cdots+n\alpha_{nn}}{n}\stackrel{?}=0.$$
This is the solution:

This is the $\epsilon$-$\delta$ solution given by a teacher, here we assume $f'(0)=1$

 A: Here's the $\varepsilon$ version basically given by your teacher although I have streamlined it a bit (at least in my mind). I provide this for future purposes in case the original image is deleted or someone searches the specific Tex.
Let $\varepsilon>0$ be given. By Taylor's Theorem, there exists a function $h(x)$ such that
$$f(x)=f(0)+f'(0)x+h(x)x=f'(0)x+h(x)x$$
and
$$\lim_{x\to 0}h(x)=0$$
Let $N_1\in\mathbb{N}$ be given such that $|h(x)|\leq \frac{\varepsilon}{2}$ whenever $|x|\leq \frac{1}{N_1}$ and let $N_2\in\mathbb{N}$ be given by
$$N_2=\left\lceil\frac{|f'(0)|}{\varepsilon}\right\rceil$$
Define $N=\max\{N_1,N_2\}$. Then for $n\geq N$ we have
$$\begin{align}\left|x_n-\frac{f'(0)}{2}\right|&=\left|\sum_{i=1}^nf\left(\frac{i}{n^2}\right)-\frac{f'(0)}{2}\right|\\&=\left|f'(0)\sum_{i=1}^n\frac{i}{n^2}+\sum_{i=1}^nh\left(\frac{i}{n^2}\right)\frac{i}{n^2}-\frac{f'(0)}{2}\right|\\&=\left|\frac{f'(0)}{2}+\frac{f'(0)}{2n}+\sum_{i=1}^nh\left(\frac{i}{n^2}\right)\frac{i}{n^2}-\frac{f'(0)}{2}\right|\\&\leq\left|\frac{f'(0)}{2n}\right|+\left|\sum_{i=1}^nh\left(\frac{i}{n^2}\right)\frac{i}{n^2}\right|\end{align}$$
Then using our definition of $N$ we have
$$\leq\frac{\varepsilon}{2}+\left|\sum_{i=1}^n\frac{\varepsilon}{2}\frac{i}{n^2}\right|=\frac{\varepsilon}{2}+\frac{\varepsilon}{2}\sum_{i=1}^n\frac{i}{n^2}<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}\sum_{i=1}^n\frac{n}{n^2}=\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$$
and the claim is proved.
