# 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!

• Your last step is wrong. The answer should come out as $0$. You can get the limit easily by using limit laws. Commented Jan 3, 2021 at 3:09
• Can you notice that each of the fractions in original expression tends to $1$ and hence the limit should be $1-1$? Commented Jan 3, 2021 at 3:10

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.

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?

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

$$\frac{n}{n+1}-\frac{n+1}{n}=\frac{n2-(n^2+2n+1)}{n(n+1)}=-\frac{n+(n+1)}{n(n+1)}=-\bigg(\frac{1}{n+1}+\frac{1}{n}\bigg)$$

$$\therefore lim_{n\rightarrow\infty} \frac{n}{n+1}-\frac{n+1}{n}=0$$