Đáp án:
\[\mathop {\lim }\limits_{x \to - \infty } \frac{{{x^2} - \left( {4 + 1} \right)x + 4}}{{{x^2} - 4}} = 1\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to - \infty } \frac{{{x^2} - \left( {4 + 1} \right)x + 4}}{{{x^2} - 4}}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{{x^2} - 5x + 4}}{{{x^2} - 4}}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{\frac{{{x^2} - 5x + 4}}{{{x^2}}}}}{{\frac{{{x^2} - 4}}{{{x^2}}}}}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{1 - \frac{5}{x} + \frac{4}{{{x^2}}}}}{{1 - \frac{4}{{{x^2}}}}}\\
= \frac{{1 - 0 + 0}}{{1 - 0}} = 1\\
\Rightarrow \mathop {\lim }\limits_{x \to - \infty } \frac{{{x^2} - \left( {4 + 1} \right)x + 4}}{{{x^2} - 4}} = 1
\end{array}\)
