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Bài tập 19 trang 226 SGK Toán 11 NC

Bài tập 19 trang 226 SGK Toán 11 NC

Tính giới hạn của các hàm số sau :

a. \(\mathop {\lim }\limits_{x \to  - 1} \frac{{{x^2} + x + 10}}{{{x^3} + 6}}\)

b. \(\mathop {\lim }\limits_{x \to  - 5} \frac{{{x^2} + 11x + 30}}{{25 - {x^2}}}\)

c. \(\mathop {\lim }\limits_{x \to  - \infty } \frac{{{x^6} + 4{x^2} + x - 2}}{{{{\left( {{x^3} + 2} \right)}^2}}}\)

d. \(\mathop {\lim }\limits_{x \to  + \infty } \frac{{{x^2} + x - 40}}{{2{x^5} + 7{x^4} + 21}}\)

e. \(\mathop {\lim }\limits_{x \to  - \infty } \frac{{\sqrt {2{x^4} + 4{x^2} + 3} }}{{2x + 1}}\)

f. \(\mathop {\lim }\limits_{x \to  + \infty } \left( {2x + 1} \right)\sqrt {\frac{{x + 1}}{{2{x^3} + x}}} \)

g. \(\mathop {\lim }\limits_{x \to  + \infty } \sqrt {9{x^2} + 11x - 100} \)

h. \(\mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {5{x^2} + 1}  - x\sqrt 5 } \right)\)

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

VDO.AI

Hướng dẫn giải chi tiết

 
 

a. \(\mathop {\lim }\limits_{x \to  - 1} \frac{{{x^2} + x + 10}}{{{x^3} + 6}} = \frac{{1 + \left( { - 1} \right) + 10}}{{ - 1 + 6}} = 2\)

b. 

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to  - 5} \frac{{{x^2} + 11x + 30}}{{25 - {x^2}}} = \mathop {\lim }\limits_{x \to  - 5} \frac{{\left( {x + 5} \right)\left( {x + 6} \right)}}{{\left( {5 - x} \right)\left( {5 + x} \right)}}\\
 = \mathop {\lim }\limits_{x \to  - 5} \frac{{x + 6}}{{5 - x}} = \frac{1}{{10}}
\end{array}\)

c. \(\mathop {\lim }\limits_{x \to  - \infty } \frac{{{x^6} + 4{x^2} + x - 2}}{{{{\left( {{x^3} + 2} \right)}^2}}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{1 + \frac{4}{{{x^4}}} + \frac{1}{{{x^5}}} - \frac{2}{{{x^6}}}}}{{{{\left( {1 + \frac{2}{{{x^3}}}} \right)}^2}}} = 1\)

d. \(\mathop {\lim }\limits_{x \to  + \infty } \frac{{{x^2} + x - 40}}{{2{x^5} + 7{x^4} + 21}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{\frac{1}{{{x^3}}} + \frac{1}{{{x^4}}} - \frac{{40}}{{{x^5}}}}}{{2 + \frac{7}{x} + \frac{{21}}{{{x^5}}}}} =  + \infty \)

e. Với mọi x < 0, ta có 

\(\frac{1}{x}\sqrt {2{x^4} + 4{x^2} + 3}  =  - \sqrt {2{x^2} + 4 + \frac{3}{{{x^2}}}} \)

Do đó :

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

f. 

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

g. 

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to  + \infty } \sqrt {9{x^2} + 11x - 100} \\
 = \mathop {\lim }\limits_{x \to  + \infty } x\sqrt {9 + \frac{{11}}{x} - \frac{{100}}{{{x^2}}}}  =  + \infty 
\end{array}\)

h. 

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {5{x^2} + 1}  - x\sqrt 5 } \right)\\
 = \mathop {\lim }\limits_{x \to  + \infty } \frac{1}{{\sqrt {5{x^2} + 1}  + x\sqrt 5 }} = 0
\end{array}\)

i.

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

-- Mod Toán 11 HỌC247

 
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