Calculate $ a_n\:=\:n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right) $

Hint

First, let $n=\frac 1x$ and compose Taylor series around $x=0$; back to $n$, get successively $$\sqrt{n^4+1}=n^2+\frac{1}{2 n^2}+O\left(\frac{1}{n^4}\right)$$ $$\sqrt{n^2+\sqrt{n^4+1}}=\sqrt{2} n+\frac{1}{4 \sqrt{2} n^3}+O\left(\frac{1}{n^5}\right)$$

$$a_n=\frac{n^3(n^2+\sqrt{n^4+1}-2n^2)}{\sqrt{n^2+\sqrt{n^4+1}}+\sqrt{2}n}$$ $$=\frac{n^3(\sqrt{n^4+1}-n^2)}{\sqrt{n^2+\sqrt{n^4+1}}+\sqrt{2}n}$$ $$=\frac{n^3}{(\sqrt{n^2+\sqrt{n^4+1}}+\sqrt{2}n)(\sqrt{n^4+1}+n^2)}$$ $$=\frac{1}{(\sqrt{1+\sqrt{1+\frac{1}{n^4}}}+\sqrt{2})(\sqrt{1+\frac{1}{n^4}}+1)}$$ Now, apply $\lim_n\to \infty$ on both sides and get the result.

Hope it helps:)

$$ \begin{align}L&=\lim_\limits{n\to \infty }\left(n^3\left(\sqrt{n^2+\sqrt{n^{4\:}+1}}-\sqrt{2}n\right)\right)\\& =\lim_\limits{n\to\infty}n^3\sqrt{n^2+\sqrt{n^4+1}}-n^4\sqrt{2}\\&=\lim_\limits{n\to\infty}\frac{\left(n^3\sqrt{n^2+\sqrt{n^4+1}}-n^4\sqrt{2}\right)\left(n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}\right)}{\left(n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}\right)}\\&=\lim_\limits{n\to\infty}\frac{-n^8+n^6\sqrt{n^4+1}}{n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}}\\&=\lim_\limits{n\to\infty}\dfrac{-n^5+n^3\sqrt{n^4+1}}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt 2}\\&=\lim_\limits{n\to\infty}\dfrac{-n^5+n^3\cdot n^2\sqrt{1+\dfrac1{n^4}}}{n\sqrt{1+\sqrt{1+\dfrac1{n^4}}}+n\sqrt2}\\&=\lim_\limits{n\to\infty}\dfrac{n^5\left[\sqrt{1+\dfrac1{n^4}}-1\right]}{n\left[\sqrt{1+\sqrt{1+\dfrac1{n^4}}}+\sqrt2\right]}\\&\boxed{\text{Let }n=\dfrac1m,\text{ so as }n\to\infty,m\to 0}\\&=\lim_\limits{m\to0}\dfrac{\sqrt{1+m^4}-1}{m^4\sqrt{1+\sqrt{1+m^4}}+\sqrt2}\\&\boxed{\text{As }x\approx 0,(1+x)^n\approx 1+nx}\\&=\lim_\limits{m\to 0}\dfrac{1+\dfrac12m^4-1}{m^4\sqrt{1+1+\dfrac12m^4}+\sqrt 2}\\&=\lim_\limits{m\to0}\dfrac{\dfrac12}{\sqrt{2+\dfrac12m^4}+\sqrt2}\\&=\lim_\limits{m\to0}\dfrac1{2\sqrt2\sqrt{1+\dfrac14m^4}+2\sqrt2}\\&=\lim_\limits{m\to0}\dfrac1{2\sqrt2\left(1+\dfrac18m^4\right)+2\sqrt2}\\&=\boxed{\dfrac1{4\sqrt2}}\end{align}$$