<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] 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?
</s>