Môn Toán Lớp 11: 1) lim(4x^3-3x^2+7) X tiến đến + vô cực 2) lim(x^4-2x^2+11) X tiến đến + vô cực 3) lim

Giải đáp:

$\begin{array}{l}
1)\mathop {\lim }\limits_{x \to  + \infty } \left( {4{x^3} – 3{x^2} + 7} \right)\\
 = \mathop {\lim }\limits_{x \to  + \infty } {x^3}\left( {4 – \dfrac{3}{x} + \dfrac{7}{{{x^3}}}} \right)\\
 =  + \infty \\
2)\mathop {\lim }\limits_{x \to  + \infty } \left( {{x^4} – 2{x^2} + 11} \right)\\
 = \mathop {\lim }\limits_{x \to  + \infty } {x^4}\left( {1 – \dfrac{2}{{{x^2}}} + \dfrac{{11}}{{{x^4}}}} \right)\\
 =  + \infty \\
3)\mathop {\lim }\limits_{x \to  + \infty } \dfrac{{5x – 2}}{{x + 7}}\\
 = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{5 – \dfrac{2}{x}}}{{1 + \dfrac{7}{x}}}\\
 = \dfrac{5}{1}\\
 = 5\\
4)\mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {{x^2} + 3x – 2}  – \sqrt {{x^2} + 5} } \right)\\
 = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{{x^2} + 3x – 2 – \left( {{x^2} + 5} \right)}}{{\sqrt {{x^2} + 3x – 2}  + \sqrt {{x^2} + 5} }}\\
 = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{3x – 7}}{{\sqrt {{x^2} + 3x – 2}  + \sqrt {{x^2} + 5} }}\\
 = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{3 – \dfrac{7}{x}}}{{\sqrt {1 + \dfrac{3}{x} – \dfrac{2}{{{x^2}}}}  + \sqrt {1 + \dfrac{5}{{{x^2}}}} }}\\
 = \dfrac{3}{{\sqrt 1  + \sqrt 1 }}\\
 = \dfrac{3}{2}\\
5)\mathop {\lim }\limits_{x \to {3^ – }} \dfrac{{5x – 1}}{{3 – x}}\\
 =  + \infty 
\end{array}$