Cant solve this Limit

Relevant wiki: L'Hopital's Rule - Basic

$\begin{aligned} L & = \lim_{x \to \infty} x^3\left(\sqrt{x^2+\sqrt{1+x^4}} - x\sqrt 2\right) \\ & = \lim_{x \to \infty} x^3\left(x\sqrt{1+\sqrt{\frac 1{x^4}+1}} - x\sqrt 2\right) \\ & = \lim_{x \to \infty} x^4\left(\sqrt{1+\sqrt{\frac 1{x^4}+1}} - \sqrt 2\right) & \small \color{#3D99F6} \text{Let }u = \frac 1{x^4} \\ & = \lim_{u \to 0} \frac {\sqrt{1+\sqrt{u+1}} - \sqrt 2}u & \small \color{#3D99F6} \text{A 0/0 cases, L'Hôpital's rule applies.} \\ & = \lim_{u \to 0} \frac {\frac 1{2\sqrt{1+\sqrt{u+1}}\left(2\sqrt{u+1}\right)}}1 & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }u. \\ & = \frac 1{2\sqrt{1+\sqrt{0+1}}\left(2\sqrt{0+1}\right)} \\ & = \frac 1{4\sqrt 2} = \frac 1{\sqrt{32}} \end{aligned}$

Therefore, $A = \boxed{32}$ .

$L = \lim_{x\to \infty} x^3\left(\sqrt{x^2+\sqrt {1+x^4}} - x\sqrt 2\right)$ Now rationalizing the function we have $\begin{aligned} L & = \lim_{x\to \infty} x^3\left(\dfrac{ \sqrt{1 + x^4} -x^2}{{\sqrt{x^2+\sqrt {1+x^4}} + x\sqrt 2}}\right)\\ & = \lim_{x\to \infty} x^4 \left(\dfrac{ \sqrt{1 +\dfrac{1}{ x^4} }-1 }{{\sqrt{1 +\sqrt {1+\dfrac{1}{x^4}}} + \sqrt 2}}\right) \\ & = \lim_{x\to \infty} x^4 \left(\dfrac{\cancel 1 + \dfrac{1}{2x^4} - \dfrac{1}{2!\cdot 2 x^8} + \cdots \cancel{-1}}{\sqrt 2 + \sqrt 2 } \right) \\ & = \dfrac{1}{4\sqrt 2}= \dfrac{1}{\sqrt{32}} \therefore A =\boxed{32}\end{aligned}$