Science:Math Exam Resources/Courses/MATH100/December 2014/Question 06 (b)/Solution 1

Multiply by the conjugate and then simplify:

$\begin{aligned} \sqrt{x^{2} + 2 x} - \sqrt{x^{2} - 2 x} & = (\sqrt{x^{2} + 2 x} - \sqrt{x^{2} - 2 x}) \cdot \frac{\sqrt{x^{2} + 2 x} + \sqrt{x^{2} - 2 x}}{\sqrt{x^{2} + 2 x} + \sqrt{x^{2} - 2 x}} \\ = \frac{x^{2} + 2 x - (x^{2} - 2 x)}{\sqrt{x^{2} + 2 x} + \sqrt{x^{2} - 2 x}} \\ = \frac{4 x}{\sqrt{x^{2} (1 + 2 / x)} + \sqrt{x^{2} (1 - 2 / x)}} \\ = \frac{4 x}{x (\sqrt{1 + 2 / x} + \sqrt{1 - 2 / x})} \\ = \frac{4}{\sqrt{1 + 2 / x} + \sqrt{1 - 2 / x}} \end{aligned}$

assuming $x > 0$ . Hence

$\begin{aligned} \lim_{x \to + \infty} (\sqrt{x^{2} + 2 x} - \sqrt{x^{2} - 2 x}) & = \lim_{x \to + \infty} \frac{4}{\sqrt{1 + 2 / x} + \sqrt{1 - 2 / x}} \\ = \frac{4}{2} = 2 \end{aligned}$