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

Find this limit: $\lim_{n\rightarrow \infty}\left(\frac{n}{n+1}-\frac{n+1}{n}\right)$

Let's see that:
\begin{align}
\frac{n}{n+1}-\frac{n+1}{n}&=\frac{n^2-(n+1)^2}{(n+1)(n)}\\&=\frac{n^2-n^2-2n-1}{n^2+n}\\&=\frac{-2n-1}{n^2+n}\\&=\frac{-\frac{2}{n}-1}{1+\frac{1}{n}} 
\end{align}
Así,
\begin{align}
\lim_{n \rightarrow \infty}\left ( \frac{n}{n+1}-\frac{n+1}{n} \right ) &=\lim_{n \rightarrow \infty} \frac{-\frac{2}{n}-1}{1+\frac{1}{n}}=\frac{-1}{1}=-1 
\end{align}
Am I correct? Is there another way to find it? I would really be very grateful if you can help me with this. Thank you very much!

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.
