Evaluate a limit using only the definition of limits

Question

I have the following expression which I need to prove using only the $|a_n - L| < \epsilon$ thing
$$\lim_{n\to\infty}\sqrt{n + 1 \over n} = 1$$
Normally I could do it but my problem is that I cant lose the root so Im stuck at this expression:
$$\frac{1 - \epsilon}{ \epsilon} < n\sqrt{n+1 \over n}$$
So I cant get to something like n < ...
Im sure there is a better way to approach it,

Any help appreciated thanks

P.S
I thought about saying that I can do square on the first equation but can I really do it? and even if so I think Im not allowed to do it in this question

Answer 1

$$\left|\;\sqrt\frac{n+1}n-1\;\right|=\frac{\sqrt{n+1}-\sqrt n}{\sqrt n}=\frac1{\sqrt n\left(\sqrt{n+1}+\sqrt n\right)}$$

Take now any $\;\epsilon>0\;$ . We want to check for which $\;n$' s we have

$$\frac1{\sqrt n\left(\sqrt{n+1}+\sqrt n\right)}<\epsilon$$

But now we can estimate :

$$\frac1{\sqrt n\left(\sqrt{n+1}+\sqrt n\right)}\le\frac1{\sqrt n\sqrt n}=\frac1n$$

So it is enough to know when

$$\frac1n<\epsilon\iff n>\frac1\epsilon$$

and we're done.

Answer 2

Hint. Prove that $\forall \varepsilon > 0\exists N \forall n>N: \sqrt{\frac{n+1}{n}} > 1 - \varepsilon$ and $\sqrt{\frac{n+1}{n}} < 1 + \varepsilon$.

Answer 3

Assuming $n, \epsilon > 0$, $$\begin{align*} \left|\sqrt{\frac{n+1}n}-1\right| &< \epsilon\\ \left|\sqrt{1+\frac1n}-1\right| &< \epsilon\\ \sqrt{1+\frac1n}-1 &< \epsilon\\ \sqrt{1+\frac1n} &< 1 + \epsilon\\ 1+\frac1n &< (1+\epsilon)^2\\ \frac1n &< (1+\epsilon)^2-1\\ n &> \frac1{(1+\epsilon)^2-1}\\ n &> \frac1{2\epsilon + \epsilon^2}\\ \end{align*}$$