Limit at Infinity

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,

$$\lfloor8L\rfloor=1$$