Bài tập 24 trang 152 SGK Toán 11 NC

a)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to  - \infty } \frac{{3{x^2} - x + 7}}{{2{x^3} - 1}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{{x^3}\left( {\frac{3}{x} - \frac{1}{{{x^2}}} + \frac{7}{{{x^3}}}} \right)}}{{{x^3}\left( {2 - \frac{1}{{{x^3}}}} \right)}}\\
 = \mathop {\lim }\limits_{x \to  - \infty } \frac{{\frac{3}{x} - \frac{1}{{{x^2}}} + \frac{7}{{{x^3}}}}}{{2 - \frac{1}{{{x^3}}}}} = \frac{0}{2} = 0
\end{array}\)

b)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to  - \infty } \frac{{2{x^4} + 7{x^3} - 15}}{{{x^4} + 1}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{{x^4}\left( {2 + \frac{7}{x} - \frac{{15}}{{{x^4}}}} \right)}}{{{x^4}\left( {1 + \frac{1}{{{x^4}}}} \right)}}\\
 = \mathop {\lim }\limits_{x \to  - \infty } \frac{{2 + \frac{7}{x} - \frac{{15}}{{{x^4}}}}}{{1 + \frac{1}{{{x^4}}}}} = 2
\end{array}\)

c)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to  + \infty } \frac{{\sqrt {{x^6} + 2} }}{{3{x^3} - 1}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{{x^3}\sqrt {1 + \frac{2}{{{x^6}}}} }}{{{x^3}\left( {3 - \frac{1}{{{x^3}}}} \right)}}\\
 = \mathop {\lim }\limits_{x \to  + \infty } \frac{{\sqrt {1 + \frac{2}{{{x^6}}}} }}{{3 - \frac{1}{{{x^3}}}}} = \frac{1}{3}
\end{array}\)

d) Với mọi x < 0, ta có:

\(\begin{array}{l}
\frac{{\sqrt {{x^6} + 2} }}{{3{x^3} - 1}} = \frac{{\left| {{x^3}} \right|\sqrt {1 + \frac{2}{{{x^6}}}} }}{{{x^3}\left( {3 - \frac{1}{{{x^3}}}} \right)}}\\
 = \frac{{ - {x^3}\sqrt {1 + \frac{2}{{{x^6}}}} }}{{{x^3}\left( {3 - \frac{1}{{{x^3}}}} \right)}} = \frac{{ - \sqrt {1 + \frac{2}{{{x^6}}}} }}{{3 - \frac{1}{{{x^3}}}}}
\end{array}\)

Do đó 

\(\mathop {\lim }\limits_{x \to  - \infty } \frac{{\sqrt {{x^6} + 2} }}{{3{x^3} - 1}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{ - \sqrt {1 + \frac{2}{{{x^6}}}} }}{{3 - \frac{1}{{{x^3}}}}} =  - \frac{1}{3}\)

-- Mod Toán 11 HỌC247