Bài tập 39 trang 166 SGK Toán 11 NC

a)

\(\mathop {\lim }\limits_{x \to  + \infty } \frac{{2{x^2} + x - 10}}{{9 - 3{x^3}}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{\frac{2}{x} + \frac{1}{{{x^2}}} - \frac{{10}}{{{x^3}}}}}{{\frac{9}{{{x^3}}} - 3}} = 0\)

b) Với mọi \(x \ne 0\), ta có:

\(\begin{array}{l}
\frac{{\sqrt {2{x^2} - 7x + 12} }}{{3\left| x \right| - 17}} = \frac{{\left| x \right|\sqrt {2 - \frac{7}{x} + \frac{{12}}{{{x^2}}}} }}{{\left| x \right|\left( {3 - \frac{{17}}{{\left| x \right|}}} \right)}}\\
 = \frac{{\sqrt {2 - \frac{7}{x} + \frac{{12}}{{{x^2}}}} }}{{3 - \frac{{17}}{{\left| x \right|}}}}
\end{array}\)

Do đó \(\mathop {\lim }\limits_{x \to  - \infty } \frac{{\sqrt {2{x^2} - 7x + 12} }}{{3\left| x \right| - 17}} = \frac{{\sqrt 2 }}{3}\)

-- Mod Toán 11 HỌC247