<s>[INST] How to evaluate $\lim\limits_{n\to\infty}(n^3)\cdot(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt{2})$? 
How to evaluate
  $\lim\limits_{n\to\infty}(n^3\cdot(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt{2}))$?

My though is, that since $\lim\limits_{n\to\infty}n^3=\infty$, the limit must be $\infty$, if the right hand side is greater than or equal to $1$. Otherwise it's a fraction, and there might be many possible solutions.
However, I'm far for being confident about the solution. What am I missing here? What's the "trick"?
 [/INST] Hint: \begin{align}\\&\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt{2}
\\\\=&\dfrac{n^2+\sqrt{n^4+1}-2n^2}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}
\\\\=&\dfrac{\sqrt{n^4+1}-n^2}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}
\\\\=&\dfrac{n^4+1-n^4}{\left(\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}\right)\cdot\left(\sqrt{n^4+1}+n^2\right)}\end{align}
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