Biết \(\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + m\;x + 2019} + x} \right) = - 3\). Giá trị của \(m\) bằng

* Hướng dẫn giải

Ta có:

\(\begin{array}{l}\mathop {\lim }\limits_{x \to  - \infty } \left( {\sqrt {{x^2} + m\;x + 2019}  + x} \right)\\ = \mathop {\lim }\limits_{x \to  - \infty } \frac{{{x^2} + m\;x + 2019 - {x^2}}}{{\sqrt {{x^2} + m\;x + 2019}  - x}}\\ = \mathop {\lim }\limits_{x \to  - \infty } \frac{{m\;x + 2019}}{{\sqrt {{x^2} + m\;x + 2019}  - x}}\\ = \mathop {\lim }\limits_{x \to  - \infty } \frac{{m\;x + 2019}}{{\left| x \right|\sqrt {1 + \frac{m}{x} + \frac{{2019}}{{{x^2}}}}  - x}}\\ = \mathop {\lim }\limits_{x \to  - \infty } \frac{{x\left( {m\; + \frac{{2019}}{x}} \right)}}{{ - x\sqrt {1 + \frac{m}{x} + \frac{{2019}}{{{x^2}}}}  - x}}\\ = \mathop {\lim }\limits_{x \to  - \infty } \frac{{m\; + \frac{{2019}}{x}}}{{ - \sqrt {1 + \frac{m}{x} + \frac{{2019}}{{{x^2}}}}  - 1}}\\ = \frac{m}{{ - 1 - 1}} =  - \frac{m}{2}\end{array}\)

Mà \(\mathop {\lim }\limits_{x \to  - \infty } \left( {\sqrt {{x^2} + m\;x + 2019}  + x} \right) =  - 3\)

\( \Rightarrow  - \frac{m}{2} =  - 3 \Leftrightarrow m = 6\)