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