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Finding $x$ as a function of $n$ and $y$ as a function of $n$

I have been trying to figure out how to solve this:

$x= \sqrt{n}$ and $y= \sqrt{n+1}$

I figured out that $x=2 \cdot y$ and wanted to figure out how to make $x$ and $y$ as functions of n, but couldn’t figure it out.
A solution that I have already done is:
$$y=\sqrt{1+\frac{1}{n}}=1+\frac{1}{2n}+O\left(\frac{1}{n^2}\right)$$
$$x=\sqrt{\frac{1}{n}+\frac{1}{n^2}}=1+\frac{1}{2\sqrt{n}}+O\left(\frac{1}{n^2}\right)$$
As you can see, the terms get pretty nasty and I am starting to wonder if there is a better way to find $x$ and $y$.
Thanks for all help in advance.

A:

Don’t be scared by the fact that the terms get nasty. Try to ignore the complexity to see what is the limiting factor. You have two terms that you need to get under control
$$\sqrt{n} = \sqrt{\frac{1}{n}} \times \sqrt{n}$$
Let’s start with the second one:
$$\sqrt{\frac{1}{n}} = \sqrt{n} – \frac{1}{2 \sqrt{n
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