Using the quadratic formula,
$$\alpha_n=\frac{(2n+1)-\sqrt{(2n+1)^2-2n(n+1)}}2$$
Simplifying the discriminant,
$$\alpha_n=\frac{(2n+1)-\sqrt{2n^2+2n+1}}2$$
Rationalising,
$$\alpha_n=\frac{(2n+1)^2-(2n^2+2n+1)}{2\left[(2n+1)+\sqrt{2n^2+2n+1}\right]}$$
$$=\frac{2n^2+2n}{2\left[(2n+1)+\sqrt{2n^2+2n+1}\right]}$$
$$=\frac{n^2+n}{(2n+1)+\sqrt{2n^2+2n+1}}$$
Therefore,
$$\frac{\alpha_n}{n}=\frac{n+1}{(2n+1)+\sqrt{2n^2+2n+1}}$$
Dividing numerator and denominator by
$$n,$$
$$\lim_{n\to\infty}\frac{\alpha_n}{n}=\frac1{2+\sqrt2}$$
Rationalising,
$$\frac1{2+\sqrt2}=\frac{2-\sqrt2}{2}=1-\frac1{\sqrt2}$$
Hence,
$$L=\lim_{n\to\infty}\frac{\alpha_n/n}{\sqrt2+\frac1n}$$
$$=\frac{1-\frac1{\sqrt2}}{\sqrt2}$$
$$=\frac1{\sqrt2}-\frac12$$
$$=\frac{\sqrt2-1}{2}$$
Now,
$$8L=4(\sqrt2-1)$$
Using
$$\sqrt2\approx1.414,$$
$$8L\approx4(0.414)$$
$$\approx1.656$$
Therefore,
$$\lfloor8L\rfloor=1$$
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