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Bài tập 44 trang 167 SGK Toán 11 NC

Bài tập 44 trang 167 SGK Toán 11 NC

Tìm các giới hạn sau:

a) \(\mathop {\lim }\limits_{x \to  - \infty } x\sqrt {\frac{{2{x^3} + x}}{{{x^5} - {x^2} + 3}}} \)

b) \(\mathop {\lim }\limits_{x \to  - \infty } \frac{{\left| x \right| + \sqrt {{x^2} + x} }}{{x + 10}}\)

c) \(\mathop {\lim }\limits_{x \to  + \infty } \frac{{\sqrt {2{x^4} + {x^2} - 1} }}{{1 - 2x}}\)

d) \(\mathop {\lim }\limits_{x \to  - \infty } \left( {\sqrt {2{x^2} + 1}  + x} \right)\)

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Hướng dẫn giải chi tiết

 
 

a) Với x < 0, ta có:

\(\begin{array}{l}
x\sqrt {\frac{{2{x^3} + x}}{{{x^5} - {x^2} + 3}}}  =  - \left| x \right|\sqrt {\frac{{2{x^3} + x}}{{{x^5} - {x^2} + 3}}} \\
 =  - \sqrt {\frac{{{x^2}\left( {2{x^3} + x} \right)}}{{{x^5} - {x^2} + 3}}}  =  - \sqrt {\frac{{2 + \frac{1}{{{x^2}}}}}{{1 - \frac{1}{{{x^3}}} + \frac{1}{{{x^5}}}}}} 
\end{array}\)

Do đó \(\mathop {\lim }\limits_{x \to  - \infty } x\sqrt {\frac{{2{x^3} + x}}{{{x^5} - {x^2} + 3}}}  =  - \sqrt 2 \)

b)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to  - \infty } \frac{{\left| x \right| + \sqrt {{x^2} + x} }}{{x + 10}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{\left| x \right| + \left| x \right|\sqrt {1 + \frac{1}{x}} }}{{x + 10}}\\
 = \mathop {\lim }\limits_{x \to  - \infty } \frac{{ - x - x\sqrt {1 + \frac{1}{x}} }}{{x + 10}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{ - 1 - \sqrt {1 + \frac{1}{x}} }}{{1 + \frac{{10}}{x}}} =  - 2
\end{array}\)

c)

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

(vì \(\mathop {\lim }\limits_{x \to  + \infty } x =  + \infty ,\mathop {\lim }\limits_{x \to  + \infty } \frac{{\sqrt {2 + \frac{1}{{{x^2}}} - \frac{1}{{{x^4}}}} }}{{\frac{1}{x} - 2}} =  - \frac{{\sqrt 2 }}{2} < 0\))

d)

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

(vì \(\mathop {\lim }\limits_{x \to  - \infty } \left( { - x - 1} \right) =  + \infty \))

-- Mod Toán 11 HỌC247

 
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