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:

\(\begin{array}{l}
1)\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^2} + 3x - 4}}{{{x^2} - 1}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x + 4} \right)\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{x + 4}}{{x + 1}} = \dfrac{{1 + 4}}{{1 + 1}} = \dfrac{5}{2}\\
2)\mathop {\lim }\limits_{x \to  + \infty } \dfrac{{ - 2{x^2} - 5x + 1}}{{2 - {x^2}}}\\
 = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{ - 2 - \dfrac{5}{x} + \dfrac{1}{{{x^2}}}}}{{\dfrac{2}{{{x^2}}} - 1}} = 2
\end{array}\)