\[\begin{split} &\lim_{n \to -\infty}\frac{1+\sqrt{2x^2-1}}{x}=\\[16pt] &=\lim_{n \to -\infty}\frac{1+\sqrt{2x^2-1}}{x}\cdot \frac{\frac{1}{|x|}}{\frac{1}{|x|}}=\\[16pt] &=\lim_{n \to -\infty}\dfrac{\dfrac{1}{|x|}+\sqrt{\dfrac{2x^2}{|x^2|}-\dfrac{1}{|x^2|}}}{\dfrac{x}{|x|}}=\\[16pt] &=\lim_{n \to -\infty}\dfrac{\dfrac{1}{|x|}+\sqrt{\dfrac{2x^2}{x^2}-\dfrac{1}{x^2}}}{-1}=\\[16pt] &=\lim_{n \to -\infty}\dfrac{\dfrac{1}{|x|}+\sqrt{2-\dfrac{1}{x^2}}}{-1}=\\[16pt] &=\dfrac{0+\sqrt{2-0}}{-1}=\\[16pt] &=-\sqrt{2} \end{split}\]