what is the limit of $\frac1{n+1}$ as $n\to\infty$?

Question

So far, I'ave used the squeeze theorem with functions $\frac1n$ and $-\frac1n$, and so got the limit $0$, but the answer is supposedly infinity... which makes little sense to me.

Answer 1

Your idea is correct

$$-\frac1n\le \frac1{n+1}\le \frac1n$$

then conclude by squeeze theorem noting that the LHS and RHS both tend to zero.

As an alternative you can also observe that

$$0\le \frac1{n+1}\le \frac1n$$

Answer 2

If you're talking about $lim_{n\rightarrow \infty}{\frac{1}{n+1}}$ try to think about how $\frac{1}{n+1}$ acts for really big values of $n$, whereas if you have $lim_{n\rightarrow 0}{\frac{1}{n+1}}$ try to compute $\frac{1}{n+1}$ when $n$ is a number really close to $0$.