Find the limit of $\frac{1}{n^2-\sqrt{n^4+4n^2+n}}$

Question

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?

Answer 1

HINT

Use

$$\frac{1}{n^2-\sqrt{n^4+4n^2+n}}\frac{n^2+\sqrt{n^4+4n^2+n}}{n^2+\sqrt{n^4+4n^2+n}}=\frac{n^2+\sqrt{n^4+4n^2+n}}{-4n^2-n}=\frac{1+\sqrt{1+4/n^2+1/n^3}}{-4-1/n}$$

In your attempt the following

$$n^2\sqrt{1+\frac{4}{n^2}+\frac{1}{n^3}}=n^2\cdot \sqrt1$$

is uncorrect, indeed by binomial series expansion

$$\sqrt{1+\frac{4}{n^2}+\frac{1}{n^3}}=1+\frac{2}{n^2}+o(1/n^2)$$

from here you can conclude by your attempt.

Answer 2

write it as $$\frac{n^2+\sqrt{n^4+4n^2+n}}{n^4-n^4-4n^2-n}$$ and this is $$\frac{n^2(1+\sqrt{1+1/n^2+1/n^3})}{n^2(-4-1/n^2)}$$

Answer 3

Hint There's a standard trick here called "multiplying by the conjugate": Multiply both the numerator and denominator of $$\frac{1}{a + \sqrt{b}}$$ by $a - \sqrt{b}$, which rationalizes the latter. Often limit problems like these are arranged so that doing so leads to a convenient cancellation.

Answer 4

$$ (n^2 + 2)^2 < n^4 + 4 n^2 + n < \left(n^2 + 2 + \frac{1}{2n} \right)^2 $$ $$ n^2 + 2 < \sqrt{n^4 + 4 n^2 + n \;} \; < \; n^2 + 2 + \frac{1}{2n} $$ $$ - 2 > n^2 -\sqrt{n^4 + 4 n^2 + n \;} \; > \; - 2 - \frac{1}{2n} $$ $$ - \frac{1}{2} < \frac{1}{n^2 -\sqrt{n^4 + 4 n^2 + n \;} } \; < \; - \frac{1}{2 + \frac{1}{2n}} $$ these are for $n > 4$

Answer 5

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.