<s>[INST] 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!
 [/INST] $$\frac{n}{n+1}-\frac{n+1}{n}=\frac{n2-(n^2+2n+1)}{n(n+1)}=-\frac{n+(n+1)}{n(n+1)}=-\bigg(\frac{1}{n+1}+\frac{1}{n}\bigg)$$
$$\therefore lim_{n\rightarrow\infty} \frac{n}{n+1}-\frac{n+1}{n}=0$$
</s>