Proving a limit with epsilon delta

Question

For the sequence $$s_n = \frac{n}{n+\sqrt n},$$ I am trying to prove that the limit as $n\to+\infty$ is $1$ with $\varepsilon-\delta$ type proof.

So i currently have that: $|f(x)-L| < \varepsilon$ if $x>\delta$

$|x/(x+\sqrt x)-1|<\varepsilon$

I dont know how to manipulate the LHS or what epsilon to choose that will help me with the proof any help is appreciated, thanks.

Answer 1

Observe that \begin{align*} \biggl|\frac{n}{n+\sqrt{n}} -1\biggr| = \biggl|\frac{\sqrt{n}}{n+\sqrt{n}}\biggr|\leq \frac{\sqrt{n}}{n}=\frac{1}{\sqrt{n}} \end{align*} Hence you have to choose an $n$ such that $\epsilon >\frac{1}{\sqrt{n}}$ happens.

Answer 2

For the case of sequences, the $\varepsilon-\delta$ type proof is as follows:

Given $\varepsilon>0$, you need to find $N\in\Bbb N$ (depending on $\varepsilon$) such that for all $n\ge N$, it holds that $|s_n-L|<\varepsilon$ (in your case, $|s_n-1|$).

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For that purpose, observe that $$|s_n-1|=\left|\frac{n}{n+\sqrt n}-1\right|=\left|\frac {n-(n+\sqrt n)}{n+\sqrt n}\right|=\frac{\sqrt n}{n+\sqrt n}=\frac{1}{\sqrt n+1}.$$

So, given $\varepsilon>0$, let $N\in\Bbb N$ such that $1+\sqrt N>\varepsilon$ (for example, $N:=\lfloor(\varepsilon-1)^2\rfloor$).

I think you can finish from here.