Biết \(\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + m\;x + 2019} + x} \right) = - 3\). Giá trị của \(m\) bằng
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Lời giải:
Báo saiTa 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 \(\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + m\;x + 2019} + x} \right) = - 3\)
\( \Rightarrow - \frac{m}{2} = - 3 \Leftrightarrow m = 6\)