<s>[INST] Find the limit of $\frac{1}{n^2-\sqrt{n^4+4n^2+n}}$ I was trying to solve the following task but I stumbled across something I do not understand:
Calculate:
$$\lim_{n \to \infty}\frac{1}{n^2-\sqrt{n^4+4n^2+n}}$$
my attempt was to factorize n^2 out of the squareroot:
$$$$
$$\lim_{n \to \infty}\frac{1}{n^2-\sqrt{n^4+4n^2+n}}$$
\begin{align}
\\a_n & = \frac{1}{n^2-\sqrt{n^4+4n^2+n}} \\
 & =\frac{1}{n^2-\sqrt{n^4\left(1+\frac{4}{n^2}+\frac{1}{n^3}\right)}} \\
 & =\frac{1}{n^2-n^2\sqrt{1+\frac{4}{n^2}+\frac{1}{n^3}}} \\  
\end{align}
Therefor, I thought that:
$$\lim_{n \to \infty}\frac{1}{n^2-n^2\sqrt{1+\frac{4}{n^2}+\frac{1}{n^3}}}$$
$$ = \lim_{n \to \infty}\frac{1}{\left(n^2-n^2\right)\sqrt{1}} = \infty$$
I also tried a different way where I got to the result of  $-\dfrac{1}{2}$.
I am not going to show that method here but it starts with using the 3rd binomial formula. Then, having the squareroot at the top of the fraction, I factorized $n^2$ and it all worked.
Why does this method like shown above not work?
I am very happy for any help.
P.S. This is not the only example where this kind of getting to a solution does not work for me. Are there cases where I am not allowed to factorize something?
 [/INST] You nearly had the answer yourself: In the line
$$\begin{align}
\\a_n & = \frac{1}{n^2-\sqrt{n^4+4n^2+n}} \\
 & =\frac{1}{n^2-\sqrt{n^4\left(1+\frac{4}{n^2}+\frac{1}{n^3}\right)}} \\
 & =\frac{1}{n^2-n^2\sqrt{1+\frac{4}{n^2}+\frac{1}{n^3}}} \\  
\end{align}$$
For large $n$ $$\sqrt{1+\frac{4}{n^2}+\frac{1}{n^3}}  \approx 1 + \frac{1}{2} (\frac{4}{n^2}+\frac{1}{n^3})$$ and $n^2$ this gives
$$\approx n^2 + \frac{1}{2} (4+\frac{1}{n})$$ so that the final result is
$$\frac{1}{n^2-n^2-2-\frac{1}{2 n}}$$ from which you can determine the desired result.
</s>