Question

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

Answer 1

$$\begin{gathered} \underset{x\to \infty }{\lim }\left[\frac{\sqrt{x^2+a^2}-\sqrt{x^2+b^2}}{\sqrt{x^2+c^2}-\sqrt{x^2+d^2}}\right] \\ \text{Rationalising the numerator and the denominator:} \\ \underset{x\to \infty }{\lim }\left[\frac{\left(\sqrt{x^2+a^2}-\sqrt{x^2+b^2}\right)}{\left(\sqrt{x^2+c^2}-\sqrt{x^2+d^2}\right)}\times \frac{\left(\sqrt{x^2+c^2}+\sqrt{x^2+d^2}\right)}{\left(\sqrt{x^2+c^2}+\sqrt{x^2+d^2}\right)}\times \frac{\left(\sqrt{x^2+a^2}+\sqrt{x^2+b^2}\right)}{\left(\sqrt{x^2+a^2}+\sqrt{x^2+b^2}\right)}\right] \\ =\underset{x\to \infty }{\lim }\left[\frac{\left(\sqrt{x^2+a^2}-\sqrt{x^2+b^2}\right)\left(\sqrt{x^2+a^2}+\sqrt{x^2+b^2}\right)\left(\sqrt{x^2+c^2}+\sqrt{x^2+d^2}\right)}{\left(\sqrt{x^2+c^2}-\sqrt{x^2+d^2}\right)\left(\sqrt{x^2+c^2}+\sqrt{x^2+d^2}\right)\left(\sqrt{x^2+a^2}+\sqrt{x^2+b^2}\right)}\right] \\ =\underset{x\to \infty }{\lim }\frac{\left(x^2+a^2\right)-\left(x^2+b^2\right)}{\left(x^2+c^2\right)-\left(x^2+d^2\right)}\times \left(\frac{\sqrt{x^2+c^2}+\sqrt{x^2+d^2}}{\sqrt{x^2+a^2}+\sqrt{x^2+b^2}}\right) \\ \underset{x\to \infty }{=\lim }\left(\frac{a^2-b^2}{c^2-d^2}\right)\left(\frac{\sqrt{x^2+c^2}+\sqrt{x^2+d^2}}{\sqrt{x^2+a^2}+\sqrt{x^2+b^2}}\right) \\ \text{Dividing the numerator and the denominator by}\mathit{\text{x}}\text{:} \\ \underset{x\to \infty }{\lim }\left(\frac{a^2-b^2}{c^2-d^2}\right)\left(\frac{\sqrt{1+{\displaystyle \frac{c^2}{x^2}}}+\sqrt{1+{\displaystyle \frac{d^2}{x^2}}}}{\sqrt{1+{\displaystyle \frac{1}{x^2}}}+\sqrt{1+{\displaystyle \frac{b^2}{x^2}}}}\right) \\ \mathrm{As}x\to \infty ,\frac{1}{x},\frac{1}{x^2}\to 0 \\ =\left(\frac{a^2-b^2}{c^2-d^2}\right)\left(\frac{\sqrt{1}+\sqrt{1}}{\sqrt{1}+\sqrt{1}}\right) \\ =\frac{a^2-b^2}{c^2-d^2} \end{gathered}$$