Question

Evaluate the limit:
$\underset{x\to \infty }{lim}\frac{\sqrt{x^2+a^2}+\sqrt{x^2+b^2}}{\sqrt{x^2+c^2}+\sqrt{x^2+d^2}}$

Answer 1

We have,
$\underset{x\to \infty }{lim}\frac{\sqrt{x^2+a^2}+\sqrt{x^2+b^2}}{\sqrt{x^2+c^2}+\sqrt{x^2+d^2}}$

It becomes $\frac{\infty }{\infty }$ form

$=\underset{x\to \infty }{lim}\frac{x\sqrt{1+\frac{a^2}{x^2}}+x\sqrt{1+\frac{b^2}{x^2}}}{x\sqrt{1+\frac{c^2}{x^2}}+\sqrt{1+\frac{d^2}{x^2}}}$

$=\underset{x\to \infty }{lim}\frac{x(\sqrt{1+\frac{a^2}{x^2}}+\sqrt{1+\frac{b^2}{x^2}})}{x(\sqrt{1+\frac{c^2}{x^2}}+\sqrt{1+\frac{d^2}{x^2}})}$

$=\underset{x\to \infty }{lim}\frac{\sqrt{1+\frac{a^2}{x^2}}+\sqrt{1+\frac{b^2}{x^2}}}{\sqrt{1+\frac{c^2}{x^2}}+\sqrt{1+\frac{d^2}{x^2}}}$

Where $x\to \infty$, then $\frac{1}{x}\to 0$

$=\frac{\sqrt{1+a^2\left(0\right)}+\sqrt{1+b^2\left(0\right)}}{\sqrt{1+c^2\left(0\right)}+\sqrt{1+d^2\left(0\right)}}$

$=\frac{1+1}{1+1}=1$

$\therefore \underset{x\to \infty }{lim}\frac{\sqrt{x^2+a^2}+\sqrt{x^2+b^2}}{\sqrt{x^2+c^2}+\sqrt{x^2+d^2}}$