Question

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

Answer 1

We have,

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

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

${lim}_{x\to \infty }\frac{x[\sqrt{1+\frac{a^2}{x^2}}+\sqrt{1+\frac{b^2}{x^2}}]}{x[\sqrt{1+\frac{c^2}{x^2}}+\sqrt{1+\frac{d^2}{x^2}}]}$

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

$\text{Taking}\text{limit}\text{and}\text{we}\text{get}$

$\Rightarrow \frac{\sqrt{1+\frac{a^2}{\infty }}+\sqrt{1+\frac{b^2}{\infty }}}{\sqrt{1+\frac{c^2}{\infty }+}\sqrt{1+\frac{d^2}{\infty }}}$

$\Rightarrow \frac{1+1}{1+1}=\frac{2}{2}=1$

$\text{Hence,}\text{this}\text{is}\text{the}\text{answer}\text{.}$