\( \lim _ { x \rightarrow + \infty } \sqrt { x ^ { 2 } - 1 } + \sqrt { x ^ { 2 } + 1 } - 2 x \)

To solve the limit \[ \lim _ { x \rightarrow + \infty } \left( \sqrt { x ^ { 2 } - 1 } + \sqrt { x ^ { 2 } + 1 } - 2 x \right), \] we can start by simplifying the expression inside the limit. As \(x\) approaches infinity, we notice that both square roots can be approximated: \[ \sqrt{x^2 - 1} = \sqrt{x^2(1 - \frac{1}{x^2})} = x\sqrt{1 - \frac{1}{x^2}} \approx x\left(1 - \frac{1}{2x^2}\right) = x - \frac{1}{2x} \] and \[ \sqrt{x^2 + 1} = \sqrt{x^2(1 + \frac{1}{x^2})} = x\sqrt{1 + \frac{1}{x^2}} \approx x\left(1 + \frac{1}{2x^2}\right) = x + \frac{1}{2x}. \] Now substituting these approximations back into the limit expression gives: \[ \sqrt{x^2 - 1} + \sqrt{x^2 + 1} \approx \left(x - \frac{1}{2x}\right) + \left(x + \frac{1}{2x}\right) = 2x. \] Therefore, \[ \sqrt{x^2 - 1} + \sqrt{x^2 + 1} - 2x \approx 2x - 2x = 0. \] To find a more precise limit, we can combine the two square roots after factoring them: \[ \lim_{x \to +\infty} \left(\sqrt{x^2 - 1} + \sqrt{x^2 + 1} - 2x\right) = \lim_{x \to +\infty} \left(\left(\sqrt{x^2 - 1} - x\right) + \left(\sqrt{x^2 + 1} - x\right)\right). \] Calculating each term separately yields: \[ \sqrt{x^2 - 1} - x \approx -\frac{1}{2x} \quad \text{and} \quad \sqrt{x^2 + 1} - x \approx \frac{1}{2x}, \] Thus, \[ \left(\sqrt{x^2 - 1} - x\right) + \left(\sqrt{x^2 + 1} - x\right) \approx -\frac{1}{2x} + \frac{1}{2x} = 0. \] Finally, taking the limit: \[ \lim_{x \to +\infty} \left(-\frac{1}{2x} + \frac{1}{2x}\right) = 0. \] Therefore, \[ \lim_{x \rightarrow + \infty } \sqrt { x ^ { 2 } - 1 } + \sqrt { x ^ { 2 } + 1 } - 2 x = 0. \] The final answer is \[ \boxed{0}. \]