Find this limit $\lim_{n\rightarrow \infty} \frac{n}{n+1}-\frac{n+1}{n}$. Am I correct?

Hint: Rewrite as follows: $$\begin{align*} \frac{n}{n+1}-\frac{n+1}{n}&=\frac{n+1-1}{n+1}-1-\frac{1}{n} \\ &=1-\frac{1}{n+1}-1-\frac{1}{n} \\ &=-\left(\frac{1}{n+1}+\frac{1}{n}\right). \end{align*}$$ As $n\to\infty$, what do $1/n$ and $1/(n+1)$ approach?

You have a mistake in:

$$ \frac{-2n-1}{n^2+n}=\frac{-\frac{2}{n}-1}{1+\frac{1}{n}} $$

It should be:

$$ \frac{-2n-1}{n^2+n}=\frac{-\frac{2}{n}-\frac{1}{n^2}}{1+\frac{1}{n}} $$

However, it would be better to take a factor of $n$:

$$ \frac{-2n-1}{n^2+n}=\frac{-2-\frac{1}{n}}{n+1} $$

Hence, you will find that the limit will be zero.

$\lim_{n\rightarrow \infty} \frac{n}{n+1}-\frac{n+1}{n} \implies \lim_{n\rightarrow \infty} \dfrac{1}{1+\frac{1}{n}}-1-\dfrac{1}{n} $ equals?