Show that $\lim\limits_{n \to +\infty}(\sin(\frac{1}{n^2})+\sin(\frac{2}{n^2})+\cdots+\sin(\frac{n}{n^2})) = \frac{1}{2}$

Start with $x-x^3/6<\sin(x)<x$; there are many nice proofs here.

Then we have $$ \sum_{k=1}^{n}(k/n^2)-(k/n^2)^3/6 <\sum_{k=1}^{n}\sin(k/n^2) < \sum_{k=1}^{n}k/n^2 $$Using the fact that $\sum_{k=1}^{n}k^p = \frac{n^{p+1}}{p+1}+\text{ lower order terms}$, we have $$ \frac{n^2+\cdots}{2 n^2}- \frac{n^4+\cdots}{24 n^6} <\sum_{k=1}^{n}\sin(k/n^2) < \frac{n^2+\cdots}{2 n^2}, $$and the result follows by squeezing. We can actually do a bit better with the sums using closed-form identites: $$ \frac{n^2+n}{2 n^2}- \frac{n^4+2n^3+n^2}{24 n^6} <\sum_{k=1}^{n}\sin(k/n^2) < \frac{n^2+n}{2 n^2} $$

For a given $x_n$, the largest argument inside of $\sin$ is $1/n$. You can bound $\sin$ on $[0,1/n]$ by below with the linear function that passes by $(0,0)$ and $(1/n,\sin(1/n)$. You can therefor bound $x_n$ by below with

$$y_n=(n\sin(1/n))(\sum_{k=1}^n \frac{k}{n^2})=\frac{\sin(1/n)(n+1)}{2}. $$

As $n \to \infty$, $y_n \to 1/2$.

Hope this helps!

If you do not know Taylor series, use the equivalent $$\sin(x)\sim x \qquad \text{when} \qquad x \text{ is small}$$

So $$x_n \sim \frac {1} {n^2}+\frac {2} {n^2}+\cdots+\frac {n-1} {n^2}+\frac {n} {n^2}=\frac {1+2+\cdots+(n-1)+n} {n^2}$$

I am sure that you can take it from here.