Giải Toán 11 SGK nâng cao Chương 4 Luyện tập (trang 167)

Tìm các giới hạn sau:

a) \(\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} + \frac{1}{{{x^2}}}} \right)\)

b) \(\mathop {\lim }\limits_{x \to  - 2} \frac{{{x^3} + 8}}{{x + 2}}\)

c) \(\mathop {\lim }\limits_{x \to 9} \frac{{3 - \sqrt x }}{{9 - x}}\)

d) \(\mathop {\lim }\limits_{x \to 0} \frac{{2 - \sqrt {4 - x} }}{x}\)

e) \(\mathop {\lim }\limits_{x \to  + \infty } \frac{{{x^4} - {x^3} + 11}}{{2x - 7}}\)

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

Hướng dẫn giải:

Câu a:

\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} + \frac{1}{{{x^2}}}} \right) = \mathop {\lim }\limits_{x \to 0} \frac{{x + 1}}{{{x^2}}} =  + \infty \) (vì \(\mathop {\lim }\limits_{x \to 0} \left( {x + 1} \right) = 1 > 0,\mathop {\lim }\limits_{x \to 0} {x^2} = 0\) và \({x^2} > 0,\forall x \ne 0\))

Câu b:

\(\mathop {\lim }\limits_{x \to  - 2} \frac{{{x^3} + 8}}{{x + 2}} = \mathop {\lim }\limits_{x \to  - 2} \frac{{\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)}}{{x + 2}} = \mathop {\lim }\limits_{x \to  - 2} \left( {{x^2} - 2x + 4} \right) = 12\)

Câu c:

\(\mathop {\lim }\limits_{x \to 9} \frac{{3 - \sqrt x }}{{9 - x}} = \mathop {\lim }\limits_{x \to 9} \frac{1}{{3 + \sqrt x }} = \frac{1}{6}\)

Câu d:

\(\mathop {\lim }\limits_{x \to 0} \frac{{2 - \sqrt {4 - x} }}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{4 - \left( {4 - x} \right)}}{{x\left( {2 + \sqrt {4 - x} } \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{1}{{2 + \sqrt {4 - x} }} = \frac{1}{4}\)

Câu e:

\(\mathop {\lim }\limits_{x \to  + \infty } \frac{{{x^4} - {x^3} + 11}}{{2x - 7}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{{x^3} - {x^2} + \frac{{11}}{x}}}{{2 - \frac{7}{x}}} =  + \infty \)

Câu f:

Với x < 0, ta có: \(\frac{{\sqrt {{x^4} + 4} }}{{x + 4}} = \frac{{{x^2}\sqrt {1 + \frac{4}{{{x^4}}}} }}{{x + 4}} = \frac{{x\sqrt {1 + \frac{4}{{{x^4}}}} }}{{1 + \frac{4}{x}}}\)

Vì \(\mathop {\lim }\limits_{x \to  - \infty } x\sqrt {1 + \frac{4}{{{x^4}}}}  =  - \infty ,\mathop {\lim }\limits_{x \to  - \infty } \left( {1 + \frac{4}{x}} \right) = 1 > 0\) nên \(\mathop {\lim }\limits_{x \to  - \infty } \frac{{\sqrt {{x^4} + 4} }}{{x + 4}} =  - \infty \)

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Tìm các giới hạn sau:

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

b) \(\mathop {\lim }\limits_{x \to 4} \frac{{\sqrt x  - 2}}{{{x^2} - 4x}}\)

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

d) \(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {{x^2} + x + 1}  - 1}}{{3x}}\)

Hướng dẫn giải:

Câu a:

Ta có \(\frac{{{x^3} + 3\sqrt 3 }}{{3 - {x^2}}} = \frac{{\left( {x + \sqrt 3 } \right)\left( {{x^2} - x\sqrt 3  + 3} \right)}}{{\left( {x + \sqrt 3 } \right)\left( {\sqrt 3  - x} \right)}} = \frac{{{x^2} - x\sqrt 3  + 3}}{{\sqrt 3  - x}}\)

Do đó \(\mathop {\lim }\limits_{x \to  - \sqrt 3 } \frac{{{x^3} + 3\sqrt 3 }}{{3 - {x^2}}} =\mathop {\lim }\limits_{x \to  - \sqrt 3 } \frac{{{x^2} - x\sqrt 3  + 3}}{{\sqrt 3  - x}} = \frac{9}{{2\sqrt 3 }} = \frac{{3\sqrt 3 }}{2}\)

Câu b:

\(\mathop {\lim }\limits_{x \to 4} \frac{{\sqrt x  - 2}}{{{x^2} - 4x}} = \mathop {\lim }\limits_{x \to 4} \frac{{\sqrt x  - 2}}{{x\left( {x - 4} \right)}} = \mathop {\lim }\limits_{x \to 4} \frac{1}{{x\left( {\sqrt x  + 2} \right)}} = \frac{1}{{16}}\)

Câu c:

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

Câu d:

\(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {{x^2} + x + 1}  - 1}}{{3x}} = \mathop {\lim }\limits_{x \to 0} \frac{{{x^2} + x + 1 - 1}}{{3x\left( {\sqrt {{x^2} + x + 1}  + 1} \right)}} = \frac{1}{3}\mathop {\lim }\limits_{x \to 0} \frac{{x + 1}}{{\sqrt {{x^2} + x + 1}  + 1}} = \frac{1}{6}\)

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

Hướng dẫn giải:

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

Câu b:

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

Câu c:

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

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

Câu d:

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

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

---

Tìm các giới hạn sau:

a) \(\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\sqrt {{x^2} + x}  - \sqrt x }}{{{x^2}}}\)

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

c) \(\mathop {\lim }\limits_{x \to {3^ - }} \frac{{3 - x}}{{\sqrt {27 - {x^3}} }}\)

d) \(\mathop {\lim }\limits_{x \to {2^ + }} \frac{{\sqrt {{x^3} - 8} }}{{{x^2} - 2x}}\)

Hướng dẫn giải:

Câu a:

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

Câu b:

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

Câu c:

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to {3^ - }} \frac{{3 - x}}{{\sqrt {27 - {x^3}} }} = \mathop {\lim }\limits_{x \to {3^ - }} \frac{{{{\left( {\sqrt {3 - x} } \right)}^2}}}{{\sqrt {\left( {3 - x} \right)\left( {{x^2} + 3x + 9} \right)} }}\\
 = \mathop {\lim }\limits_{x \to {3^ - }} \frac{{\sqrt {3 - x} }}{{\sqrt {{x^2} + 3x + 9} }} = 0
\end{array}\)

Câu d:

\(\mathop {\lim }\limits_{x \to {2^ + }} \frac{{\sqrt {{x^3} - 8} }}{{{x^2} - 2x}} = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{\sqrt {\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)} }}{{x\left( {x - 2} \right)}} = \mathop {\lim }\limits_{x \to {2^ + }} \frac{1}{x}\sqrt {\frac{{{x^2} + 2x + 4}}{{x - 2}}}  =  + \infty \)

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

 

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