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 $\dfrac{1}{4\sqrt{2}}$.
 A: HINT
You only need to apply the trick twice, indeed we have that
$$\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt{2}=(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt{2})\cdot\frac{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}=$$$$=\frac{n^2+\sqrt{n^4+1}-2n^2}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}$$
and
$$\frac{\sqrt{n^4+1}-n^2}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}=\frac{\sqrt{n^4+1}-n^2}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}\cdot \frac{\sqrt{n^4+1}+n^2}{\sqrt{n^4+1}+n^2}=$$$$=\frac{1}{(\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2})(\sqrt{n^4+1}+n^2)}$$
Can you conclude form here?
A: Let $1/n=h$
$$\lim_{h\to0^+}\dfrac{\sqrt{1+\sqrt{1+h^4}}-\sqrt2}{h^4}$$
$$=\lim_{h\to0^+}\dfrac{1+\sqrt{1+h^4}-2}{h^4}\cdot\lim_{h\to0^+}\dfrac1{\sqrt{1+\sqrt{1+h^4}}+\sqrt2}$$
$$=\lim_{h\to0^+}\dfrac{1+h^4-1}{h^4}\cdot\lim_{h\to0^+}\dfrac1{\sqrt{1+h^4}+1}\cdot\lim_{h\to0^+}\dfrac1{\sqrt{1+\sqrt{1+h^4}}+\sqrt2}$$
$$=\dfrac1{(\sqrt1+1)(\sqrt{1+\sqrt1}+\sqrt2)}$$
A: The expedite way:
$$\sqrt{1+\sqrt{1+n^{-4}}}=\sqrt{1+1+\dfrac12n^{-4}+o(n^{-4})}=\sqrt2\sqrt{1+\dfrac14n^{-4}+o(n^{-4})}=\sqrt2\left(1+\dfrac18n^{-4}+o(n^{-4})\right)$$
and the limit is
$$\frac{\sqrt2}8.$$
A: First replace $1/x=h$ to find 
$$L=\lim_{n→∞}n^3(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt2)=\lim_{h\to0^+}\dfrac{\sqrt{1+\sqrt{1+h^4}}-\sqrt2}{h^4}$$
Let $\sqrt{1+\sqrt{1+h^4}}=y+\sqrt2\implies1+\sqrt{1+h^4}=(\sqrt2+y)^2=2+2\sqrt2y+y^2$
$$1+h^4=(1+y(2\sqrt2+y))^2=1+2y(2\sqrt2+y)+y^2(2\sqrt2+y)^2$$
$$L=\lim_{y\to0}\dfrac{1+2y(2\sqrt2+y)+y^2(2\sqrt2+y)^2-1}y=?$$
A: 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 $$f'(0)=\frac{1}{4\sqrt{2}}$$

