Biết rằng $\lim_{x \to -\infty} \Big( \sqrt{5x^{2} + 2x} + x\sqrt{5} \Big) = a\sqrt{5} + b$, với $a, b \in \mathbb{Q}$. Tính $S = 5a + b$.

Lời giải Ta có: $$ \lim_{x \to -\infty} \Big( \sqrt{5x^{2} + 2x} + x\sqrt{5} \Big) = \lim_{x \to -\infty} \frac{2x}{\sqrt{5x^{2} + 2x} - x\sqrt{5}} $$ $$ = \lim_{x \to -\infty} \frac{2}{-\sqrt{5 + \frac{2}{x}} - \sqrt{5}} = \frac{2}{-2\sqrt{5}} = -\frac{1}{\sqrt{5}} = -\frac{1}{5}\sqrt{5} $$ $$ = a\sqrt{5} + b \Rightarrow \left\{\begin{array}{l} a = -\frac{1}{5} \\ b = 0 \end{array}\right. \Rightarrow S = 5a + b = -1 $$