$\underset{x\to -\infty }{\mathop{\lim }}\,\frac{x-\sqrt{{{x}^{2}}+1}}{\left| x \right|-\sin x}=\underset{x\to -\infty }{\mathop{\lim }}\,\frac{x-\left| x \right|}{\left| x \right|-\sin x}$  چون همواره $-1\le \sin x\le 1$، بنابراین:  $=\underset{x\to -\infty }{\mathop{\lim }}\,\frac{x-\left| x \right|}{\left| x \right|}=\underset{x\to -\infty }{\mathop{\lim }}\,\frac{x-(-x)}{-x}=\underset{x\to -\infty }{\mathop{\lim }}\,\frac{2x}{-x}=-2$