$\text{ If }\underset{x\to \infty }{\lim }{x}^{\alpha }\left(\sqrt{x^2+\sqrt{x^4+1}}-\sqrt{2}x\right)\text{ exists and has value non-zero finite real number }$$L,\text{ then the value of }100-\frac{\alpha }{L^2}\text{ is }$

$\begin{array}{r}\text{ we have }\underset{x\to \mathrm{\infty }}{lim} x^{\alpha }\frac{\left(x^2+\sqrt{x^4+1}\right)-2x^2}{{\left(x^2+\sqrt{x^4+1}\right)}^{1/2}+\sqrt{2}x}\\ =\underset{x\to \mathrm{\infty }}{lim} x^{\alpha }\frac{\left(\sqrt{x^4+1}-x^2\right)}{\left({\left(x^2+\sqrt{x^4+1}\right)}^{1/2}+\sqrt{2}x\right)}\\ =\underset{x\to \mathrm{\infty }}{lim} x^{\alpha +1}\frac{\left(\sqrt{1+\frac{1}{x^4}}-1\right)}{{\left(\sqrt{1+\frac{1}{x^4}}+1\right)}^{1/2}+\sqrt{2}}\\ =\frac{1}{2\sqrt{2}}\underset{x\to \mathrm{\infty }}{lim} x^{\alpha +1}\left(1+\frac{1}{2}\cdot \frac{1}{x^4}+\dots -1\right)\end{array}$

So, for above limit to exist and has value non-zero, we must have

$\begin{array}{l}\alpha +1=4\Rightarrow \alpha =3\text{ and }L=\frac{1}{2\sqrt{2}}\times \frac{1}{2}=\frac{1}{4\sqrt{2}}\\ \Rightarrow L^2=\frac{1}{32}\end{array}$