Limit of a sum 1: $\lim _{n\to \infty }\sum _{k=1}^n\sqrt{n^4+k} \cdot \sin \frac{2k\pi }{n}$ $$\lim _{n\to \infty }\sum _{k=1}^n\sqrt{n^4+k} \cdot \sin \frac{2k\pi }{n}$$
It looks like a Riemann sum, but I don't know how to approach it. Any hint would be appreciated. (Without Taylor expansion)
EDIT: The answer is $- \frac{1}{4 \pi}$

I don't have time to finish right now; I'll leave this as Community Answer, so that anyone can feel free to edit it; if no one does, I'll get back to it later.
The inhomogeneity between $n^4$ and $k$ will a priori prevent you from seeing this as a Riemann sum: since there is no way to massage $\sqrt{n^4+k}$ in order to make a factor $\sqrt{1+\frac{k}{n}}$ appear.
However, what you can do is "guess and then use inequalities and the squeeze theorem to prove the guess was right." I know you do not want to use Taylor expansions, so I will not (but it's hidden in the "guess" part).
First, rewrite
$$
\sum_{k=1}^n \sqrt{n^4+k}\sin\frac{2\pi k}{n}
= n^3 \frac{1}{n}\sum_{k=1}^n \sqrt{1+\frac{k}{n^4}}\sin\frac{2\pi k}{n}
$$


*

*Guess part: Now, we are going to use the fact that 
$$\lim_{x\to0} \frac{\sqrt{1+x}-1}{x} = \lim_{x\to0} \frac{1}{\sqrt{1+x}+1} = \frac{1}{2}\tag{i}$$, so we expect $\sqrt{1+\frac{k}{n^4}}-1$ to "behave like $\frac{k}{2n^4}$" and (ii) $$\frac{1}{n}\sum_{k=1}^n\sin\frac{2\pi k}{n} \xrightarrow[n\to\infty]{} \int_0^1 \sin(2\pi x)\, dx = 0 \tag{ii}$$
(Riemann sum).
Combining the two, it is "reasonable" to expect that
$$\begin{align}
n^3 \frac{1}{n}\sum_{k=1}^n \sqrt{1+\frac{k}{n^4}}\sin\frac{2\pi k}{n}
&= 
n^3 \frac{1}{n}\sum_{k=1}^n \left(\sqrt{1+\frac{k}{n^4}}-1\right)\sin\frac{2\pi k}{n} + n^3\cdot\frac{1}{n}\sum_{k=1}^n\sin\frac{2\pi k}{n} \\
&\approx
n^3 \frac{1}{n}\sum_{k=1}^n \frac{k}{2n^4}\sin\frac{2\pi k}{n} + n^3\cdot\frac{1}{n}\sum_{k=1}^n\sin\frac{2\pi k}{n} \\
&= \frac{1}{2}\cdot\frac{1}{n}\sum_{k=1}^n \frac{k}{n}\sin\frac{2\pi k}{n} + n^3\cdot\frac{1}{n}\sum_{k=1}^n\sin\frac{2\pi k}{n} \\
&\xrightarrow[n\to\infty]{} \frac{1}{2}\int_0^1 x\sin(2\pi x)\, dx+\frac{1}{2}\int_0^1 \sin(2\pi x)\, dx =  \frac{1}{2}\int_0^1 x\sin(2\pi x)\, dx \\&= \boxed{-\frac{1}{4\pi}}
\end{align}$$
This is our (very handwavy) guess; let us prove it.

*Proving the guess:
