Bài 42 trang 167 SGK Toán 11 nâng cao
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 \(\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {{x^4} + 4} }}{{x + 4}} = - \infty \)
Bài 43 trang 167 SGK Toán 11 nâng cao
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:
\(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {{x^2} + x + 1} - 1}}{{3x}} = \mathop {\lim }\limits_{x \to 0} \frac{{{x^2} + x + 1 - 1}}{{3x\left( {\sqrt {{x^2} + x + 1} + 1} \right)}} = \frac{1}{3}\mathop {\lim }\limits_{x \to 0} \frac{{x + 1}}{{\sqrt {{x^2} + x + 1} + 1}} = \frac{1}{6}\)
Bài 44 trang 167 SGK Toán 11 nâng cao
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 \(\mathop {\lim }\limits_{x \to - \infty } \left( { - x - 1} \right) = + \infty \))
Bài 45 trang 167 SGK Toán 11 nâng cao
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 \(x\sqrt {x - 2} > 0,\forall x > 2\)).
Trên đây là nội dung chi tiết Giải bài tập nâng cao Toán 11 Chương 4 Luyện tập (trang 167) với hướng dẫn giải chi tiết, rõ ràng, trình bày khoa học. Chúng tôi hy vọng đây sẽ là tài liệu hữu ích giúp các bạn học sinh lớp 11 học tập thật tốt.
