<s>[INST] Finding $\lim\limits_{n→∞}n^3(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt2)$ What is$$\lim_{n→∞}n^3(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt2)?$$So it is$$\lim_{n→∞}\frac{n^3(\sqrt{n^2+\sqrt{n^4+1}})^2-(n\sqrt{2})^2}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}=\lim_{n→∞}\frac{n^3(n^2+\sqrt{n^4+1}-2n^2)}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}.$$
I do not know what to do next, because my resuts is $∞$ but the answer from book is $\dfrac{1}{4\sqrt{2}}$.
 [/INST] Formally substitute $n=1/t$; if the function you get has a limit for $t\to0^+$, then it is the same as the limit you are looking for. So consider
$$
\lim_{t\to0^+}\frac{1}{t^3}\left(
  \sqrt{\frac{1}{t^2}+\sqrt{\frac{1}{t^4}+1}}-\frac{\sqrt{2}}{t}
\right)=
\lim_{t\to0^+}\frac{\sqrt{1+\sqrt{1+t^4}}-\sqrt{2}}{t^4}
$$
Now the dependency is only on $t^4$, so the limit is the same as
$$
\lim_{u\to0^+}\frac{\sqrt{1+\sqrt{1+u}}-\sqrt{2}}{u}
$$
which is the derivative at $0$ of $f(u)=\sqrt{1+\sqrt{1+u}}$.

 Since $$f'(u)=\frac{1}{2\sqrt{1+\sqrt{1+u}}}\frac{1}{2\sqrt{1+u}}$$ we have $$f'(0)=\frac{1}{4\sqrt{2}}$$

</s>