1. tìm lim (x^2 + 3x – 4) / (x^2 – 1) 02/10/2021 Bởi Margaret 1. tìm lim (x^2 + 3x – 4) / (x^2 – 1) x–>1 2. tìm lim (-2x^2 – 5x + 1) / (2 – x^2) x–>+vô cực
Đáp án: \(1)\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^2} + 3x – 4}}{{{x^2} – 1}} = \dfrac{5}{2}\) Giải thích các bước giải: \(\begin{array}{l}1)\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^2} + 3x – 4}}{{{x^2} – 1}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x + 4} \right)\left( {x – 1} \right)}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{x + 4}}{{x + 1}} = \dfrac{{1 + 4}}{{1 + 1}} = \dfrac{5}{2}\\2)\mathop {\lim }\limits_{x \to + \infty } \dfrac{{ – 2{x^2} – 5x + 1}}{{2 – {x^2}}}\\ = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{ – 2 – \dfrac{5}{x} + \dfrac{1}{{{x^2}}}}}{{\dfrac{2}{{{x^2}}} – 1}} = 2\end{array}\) Bình luận
Đáp án:
\(1)\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^2} + 3x – 4}}{{{x^2} – 1}} = \dfrac{5}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^2} + 3x – 4}}{{{x^2} – 1}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x + 4} \right)\left( {x – 1} \right)}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{x + 4}}{{x + 1}} = \dfrac{{1 + 4}}{{1 + 1}} = \dfrac{5}{2}\\
2)\mathop {\lim }\limits_{x \to + \infty } \dfrac{{ – 2{x^2} – 5x + 1}}{{2 – {x^2}}}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{ – 2 – \dfrac{5}{x} + \dfrac{1}{{{x^2}}}}}{{\dfrac{2}{{{x^2}}} – 1}} = 2
\end{array}\)
Đáp án:
Giải thích các bước giải: