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

calculuslimitssequences-and-series

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!

Best Answer

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.

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