$\underset{x\to +\infty }{\mathop{\lim }}\,\frac{ax-\sqrt{{{x}^{2}}+x}}{3x-1}=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{ax-\left| x+\frac{1}{2} \right|}{3x-1}=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{(a-1)x-\frac{1}{2}}{3x-1}=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{(a-1)x}{3x}=\frac{a-1}{3}=2\Rightarrow a=7$ بنابراین: $\underset{x\to -\infty }{\mathop{\lim }}\,f(x)=\underset{x\to -\infty }{\mathop{\lim }}\,\frac{7x-\sqrt{{{x}^{2}}+x}}{3x-1}=\underset{x\to -\infty }{\mathop{\lim }}\,\frac{7x-\left| x \right|}{3x}=\underset{x\to -\infty }{\mathop{\lim }}\,\frac{7x+x}{3x}=\underset{x\to -\infty }{\mathop{\lim }}\,\frac{8x}{3x}=\frac{8}{3}$