Question

Solve ${lim}_{x\to \infty }(\sqrt{x^2+ax+a^2}-\sqrt{x^2+a^2})$ =

Answer 1

${lim}_{x\to \infty }(\sqrt{x^2+ax+a^2}-\sqrt{x^2+a^2})$

$={lim}_{x\to \infty }\left(\frac{(\sqrt{x^2+ax+a^2}{)}^2-(\sqrt{x^2+a^2}{)}^2}{\sqrt{x^2+ax+a^2}+\sqrt{x^2+a^2}}\right)$

$={lim}_{x\to \infty }\left(\frac{x^2+ax+a^2-x^2-a^2}{\sqrt{x^2+ax+a^2}+\sqrt{x^2+a^2}}\right)$

$={lim}_{x\to \infty }\left(\frac{ax}{\sqrt{x^2+ax+a^2}+\sqrt{x^2+a^2}}\right)$

$={lim}_{x\to \infty }\left(\frac{ax}{x\sqrt{1+\frac{a}{x}+\frac{a^2}{x^2}}+x\sqrt{1+\frac{a^2}{x^2}}}\right)$

$={lim}_{x\to \infty }\left(\frac{a}{\sqrt{1+\frac{a}{x}+\frac{a^2}{x^2}}+\sqrt{1+\frac{a^2}{x^2}}}\right)$

as $x\to \infty$ then $\frac{1}{x}\to 0$

$={lim}_{\frac{1}{x}\to 0}\left(\frac{a}{\sqrt{1+\frac{a}{x}+\frac{a^2}{x^2}}+\sqrt{1+\frac{a^2}{x^2}}}\right)$

$=\frac{a}{\sqrt{1+0+0}+\sqrt{1+0}}$

$=\frac{a}{\sqrt{2}}$