Bài tập 34 trang 163 SGK Toán 11 NC

Bài tập 34 trang 163 SGK Toán 11 NC

a)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to  - \infty } \left( {3{x^3} - 5{x^2} + 7} \right)\\
 = \mathop {\lim }\limits_{x \to  - \infty } {x^3}\left( {3 - \frac{5}{x} + \frac{7}{{{x^3}}}} \right) =  - \infty 
\end{array}\)

(vì \(\mathop {\lim }\limits_{x \to  - \infty } {x^3} =  - \infty ,\mathop {\lim }\limits_{x \to  - \infty } \left( {3 - \frac{5}{x} + \frac{7}{{{x^3}}}} \right) = 3 > 0\))

b)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to  + \infty } \sqrt {2{x^4} - 3x + 12} \\
 = \mathop {\lim }\limits_{x \to  + \infty } {x^2}\sqrt {2 - \frac{3}{{{x^3}}} + \frac{{12}}{{{x^4}}}}  =  + \infty 
\end{array}\)

(vì \(\mathop {\lim }\limits_{x \to  + \infty } {x^2} =  + \infty ,\mathop {\lim }\limits_{x \to  + \infty } \sqrt {2 - \frac{3}{{{x^3}}} + \frac{{12}}{{{x^4}}}}  = \sqrt 2  > 0\)).

-- Mod Toán 11