Bài tập 40 trang 166 SGK Toán 11 NC

a) Với x > - 1 đủ gần - 1 (- 1 < x < 0), ta có:

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\left( {{x^3} + 1} \right)\sqrt {\frac{x}{{{x^2} - 1}}} \\
 = \left( {{x^2} - x + 1} \right)\left( {x + 1} \right)\sqrt {\frac{x}{{{x^2} - 1}}} 
\end{array}\\
{ = \left( {{x^2} - x + 1} \right)\sqrt {\frac{{x\left( {x + 1} \right)}}{{x - 1}}} }\\
\begin{array}{l}
 \Rightarrow \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \left( {{x^3} + 1} \right)\sqrt {\frac{x}{{{x^2} - 1}}} \\
 = \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \left( {{x^2} - x + 1} \right)\sqrt {\frac{{x\left( {x + 1} \right)}}{{x - 1}}}  = 0
\end{array}
\end{array}\)

b)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\mathop {\lim }\limits_{x \to  + \infty } \left( {x + 2} \right)\sqrt {\frac{{x - 1}}{{{x^3} + x}}} \\
 = \mathop {\lim }\limits_{x \to  + \infty } \sqrt {\frac{{{{\left( {x + 2} \right)}^2}\left( {x - 1} \right)}}{{{x^3} + x}}} 
\end{array}\\
{ = \mathop {\lim }\limits_{x \to  + \infty } \sqrt {\frac{{{{\left( {1 + \frac{2}{x}} \right)}^2}\left( {1 - \frac{1}{x}} \right)}}{{1 + \frac{1}{{{x^2}}}}}}  = 1}
\end{array}\)

-- Mod Toán 11 HỌC247