For what value of $α$ the limit $\underset{x→∞}{\lim}\sqrt{2α^2x^2+αx+7}-\sqrt{2α^2x^2+7}$ will be $\frac{1}{2\sqrt{2}}$

$α≠0$

$\underset{x→∞}{\lim}\sqrt{2α^2x^2+αx+7}-\sqrt{2α^2x^2+7}=\frac{1}{2\sqrt{2}}$

$⇒\underset{x→∞}{\lim}\frac{(2α^2x^2+αx+7)-(2α^2x^2+7)}{\sqrt{2α^2x^2+αx+7}+\sqrt{2α^2x^2+7}}=\frac{1}{2\sqrt{2}}$

$⇒\underset{x→∞}{\lim}\frac{αx}{x\left(\sqrt{2α^2+\frac{α}{x}+\frac{7}{x^2}}+\sqrt{2α^2+\frac{7}{x^2}}\right)}=\frac{1}{2\sqrt{2}}$

$⇒\frac{α}{2\sqrt{2}α}=\frac{1}{2\sqrt{2}}$ this is possible when $α≠0$ (most correct)