Đáp án:
\[\mathop {\lim }\limits_{x \to - \infty } \left( {4x - \sqrt {16{x^2} + 5} } \right) = - \infty \]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to - \infty } \left( {4x - \sqrt {16{x^2} + 5} } \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \left( {4x - \sqrt {{x^2}.\left( {16 + \frac{5}{{{x^2}}}} \right)} } \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \left( {4x - \left| x \right|.\sqrt {16 + \frac{5}{{{x^2}}}} } \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \left( {4x - \left( { - x} \right).\sqrt {16 + \frac{5}{{{x^2}}}} } \right)\,\,\,\,\,\,\,\,\,\,\,\left( {x \to - \infty \Rightarrow x < 0 \Rightarrow \left| x \right| = - x} \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \left( {4x + x.\sqrt {16 + \frac{5}{{{x^2}}}} } \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \left[ {x.\left( {4 + \sqrt {16 + \frac{5}{{{x^2}}}} } \right)} \right]\\
= - \infty \\
\left( \begin{array}{l}
\mathop {\lim }\limits_{x \to - \infty } x = - \infty \\
\mathop {\lim }\limits_{x \to - \infty } \left( {4 + \sqrt {16 + \frac{5}{{{x^2}}}} } \right) = 4 + \sqrt {16} = 8
\end{array} \right)
\end{array}\)
