Need help with limit proving. Is it safe to say that $\frac{\sqrt{n+1}}{\sqrt{n}}\rightarrow1$ if $$\lim_{n\to\infty}(\sqrt{n+1}-\sqrt{n})=0$$?
Because I want to prove that $\sqrt{n+1}\sim\sqrt{n}\ $when $n\to\infty$, but I don't know how to approach from $\frac{\sqrt{n+1}}{\sqrt{n}}\rightarrow1$. So I was thinking whether by showing $\lim_{n\to\infty}(\sqrt{n+1}-\sqrt{n})=0$ will work.
Thx guys.
Edited: I forgot to mention I have to use the $\mathcal{E}-N$ argument to prove this limit.
 A: For, $\frac{\sqrt(n+1)}{\sqrt(n)}=\sqrt(1+\frac{1}{n}) $
$\lim_{n\to\infty}\frac{\sqrt(n+1)}{\sqrt(n)}=\lim_{n\to\infty} (\sqrt(1+\frac{1}{n})) = 1 $
Also,$\sqrt(n+1)-\sqrt(n) = \frac{(\sqrt(n+1)-\sqrt(n))(\sqrt(n+1)+\sqrt(n))}{\sqrt(n+1)+\sqrt(n)}=\frac{1}{\sqrt(n+1)+\sqrt(n)} $
So,$\lim_{n\to\infty} (\sqrt(n+1)-\sqrt(n)) = \lim_{n\to\infty} (\frac{1}{\sqrt(n+1)+\sqrt(n)}) = 0 $
Then, dividing bothsides by $\sqrt(n) $ ,
So,$\lim_{n\to\infty} (\frac{\sqrt(n+1)}{\sqrt(n)}-1)=0 \implies \lim_{n\to\infty} (\sqrt(1+\frac{1}{n})) = 1 $
Edit: clearly, $\frac{\sqrt(n+1)}{\sqrt(n)}=\sqrt(1+\frac{1}{n}) $
Now by Archimedean property, for any $\epsilon>0 $ , there exists $m\in \mathbb{N} $, $\frac{1}{n}< \epsilon $ for all $n \ge m $.
Hence , $\sqrt(1+\frac{1}{n})-1 < \sqrt(1+\epsilon)-1 < (1+\epsilon)-1= \epsilon $ ,
for all $ n \ge m $
So, $\sqrt(1+\frac{1}{n})-1 < \epsilon $, for all $ n \ge m $.
A: $$L=\lim_{n\to \infty} [\sqrt{n+1}-\sqrt{n}]=\lim_{n\to \infty} \frac{(\sqrt{n+1}+\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=\lim_{n\to \infty} \frac{1}{\sqrt{n+1}+\sqrt{n}}=0$$
And if $n$ is very large, the dominant term in the numerator of $\frac{\sqrt{n+1}}{\sqrt{n}}$ is $\sqrt{n}$ and that in the denominator is $\sqrt{n}$, hence $$\frac{\sqrt{n+1}}{\sqrt{n}}\to \frac{\sqrt{n}}{\sqrt{n}}=1.$$
