1) lim (1/x-2 - 12/x ^3 -8 ) X→2. 2) lim x^3 + 3x^2 / x^3 +27 X→-3 3) lim 2x^3+5x^2+4x+1 / x^3 + x^2 -x-1 X→1 4) lim 2x^3-3x+1/x^3-1 X→-1 Giúp em các câu này vs ạh

Đáp án:

$\begin{array}{l}
1)\mathop {\lim }\limits_{x \to 2} \left( {\dfrac{1}{{x - 2}} - \dfrac{{12}}{{{x^3} - 8}}} \right)\\
 = \mathop {\lim }\limits_{x \to 2} \dfrac{{{x^2} + 2x + 4 - 12}}{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}}\\
 = \mathop {\lim }\limits_{x \to 2} \dfrac{{{x^2} + 2x - 8}}{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}}\\
 = \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {x - 2} \right)\left( {x + 4} \right)}}{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}}\\
 = \mathop {\lim }\limits_{x \to 2} \dfrac{{x + 4}}{{{x^2} + 2x + 4}}\\
 = \dfrac{{2 + 4}}{{{2^2} + 2.2 + 4}} = \dfrac{1}{2}\\
3)\mathop {\lim }\limits_{x \to 1} \dfrac{{2{x^3} + 5{x^2} + 4x + 1}}{{{x^3} + {x^2} - x - 1}}\\
 = \dfrac{{2.1 + 5.1 + 4.1 + 1}}{{{0^ + }}} = \infty \\
4)\mathop {\lim }\limits_{x \to  - 1} \dfrac{{2{x^3} - 3x + 1}}{{{x^3} - 1}}\\
 = \dfrac{{2.{{\left( { - 1} \right)}^3} - 3.\left( { - 1} \right) + 1}}{{{{\left( { - 1} \right)}^3} - 1}}\\
 = \dfrac{{ - 2 + 3 + 1}}{{ - 2}} =  - 1
\end{array}$