Bài 30 trang 159 SGK Toán 11 nâng cao
Tìm các giới hạn sau:
a) \(\mathop {\lim }\limits_{x \to \sqrt 3 } \left| {{x^2} - 8} \right|\)
b) \(\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} + x + 1}}{{{x^2} + 2x}}\)
c) \(\mathop {\lim }\limits_{x \to - 1} \sqrt {\frac{{{x^3}}}{{{x^2} - 3}}} \)
d) \(\mathop {\lim }\limits_{x \to 3} \sqrt[3]{{\frac{{2x\left( {x + 1} \right)}}{{{x^2} - 6}}}}\)
e) \(\mathop {\lim }\limits_{x \to - 2} \frac{{\sqrt {1 - {x^3}} - 3x}}{{2{x^2} + x - 3}}\)
f) \(\mathop {\lim }\limits_{x \to - 2} \frac{{2\left| {x + 1} \right| - 5\sqrt {{x^2} - 3} }}{{2x + 3}}\)
Hướng dẫn giải:
Câu a:
\(\mathop {\lim }\limits_{x \to \sqrt 3 } \left| {{x^2} - 8} \right| = \left| {{{\left( {\sqrt 3 } \right)}^2} - 8} \right| = 5\)
Câu b:
\(\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} + x + 1}}{{{x^2} + 2x}} = \frac{{{2^2} + 2 + 1}}{{{2^2} + 2.2}} = \frac{7}{8}\)
Câu c:
\(\mathop {\lim }\limits_{x \to - 1} \sqrt {\frac{{{x^3}}}{{{x^2} - 3}}} = \sqrt {\frac{1}{2}} = \frac{{\sqrt 2 }}{2}\)
Câu d:
\(\mathop {\lim }\limits_{x \to 3} \sqrt[3]{{\frac{{2x\left( {x + 1} \right)}}{{{x^2} - 6}}}} = \sqrt[3]{{\frac{{24}}{3}}} = 2\)
Câu e:
\(\mathop {\lim }\limits_{x \to - 2} \frac{{\sqrt {1 - {x^3}} - 3x}}{{2{x^2} + x - 3}} = \frac{{3 + 6}}{{8 - 5}} = 3\)
Câu f:
\(\mathop {\lim }\limits_{x \to - 2} \frac{{2\left| {x + 1} \right| - 5\sqrt {{x^2} - 3} }}{{2x + 3}} = \frac{{2 - 5}}{{ - 4 + 3}} = 3\)
Bài 31 trang 159 SGK Toán 11 nâng cao
Tìm các giới hạn sau:
a) \(\mathop {\lim }\limits_{x \to - \sqrt 2 } \frac{{{x^3} + 2\sqrt 2 }}{{{x^2} - 2}}\)
b) \(\mathop {\lim }\limits_{x \to 3} \frac{{{x^4} - 27x}}{{2{x^2} - 3x - 9}}\)
c) \(\mathop {\lim }\limits_{x \to - 2} \frac{{{x^4} - 16}}{{{x^2} + 6x + 8}}\)
d) \(\mathop {\lim }\limits_{x \to {1^ - }} \frac{{\sqrt {1 - x} + x - 1}}{{\sqrt {{x^2} - {x^3}} }}\)
Hướng dẫn giải:
Câu a:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to - \sqrt 2 } \frac{{{x^3} + 2\sqrt 2 }}{{{x^2} - 2}} = \mathop {\lim }\limits_{x \to - \sqrt 2 } \frac{{{x^3} + {{\left( {\sqrt 2 } \right)}^3}}}{{{x^2} - {{\left( {\sqrt 2 } \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to - \sqrt 2 } \frac{{\left( {x + \sqrt 2 } \right)\left( {{x^2} - x\sqrt 2 + 2} \right)}}{{\left( {x + \sqrt 2 } \right)\left( {x - \sqrt 2 } \right)}}\\
= \mathop {\lim }\limits_{x \to - \sqrt 2 } \frac{{{x^2} - x\sqrt 2 + 2}}{{x - \sqrt 2 }} = \frac{{ - 3\sqrt 2 }}{2}
\end{array}\)
Câu b:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 3} \frac{{{x^4} - 27x}}{{2{x^2} - 3x - 9}} = \mathop {\lim }\limits_{x \to 3} \frac{{x\left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right)}}{{\left( {x - 3} \right)\left( {2x + 3} \right)}}\\
= \mathop {\lim }\limits_{x \to 3} \frac{{x\left( {{x^2} + 3x + 9} \right)}}{{2x + 3}} = 9
\end{array}\)
Câu c:
\(\mathop {\lim }\limits_{x \to - 2} \frac{{{x^4} - 16}}{{{x^2} + 6x + 8}} = \mathop {\lim }\limits_{x \to - 2} \frac{{\left( {{x^2} + 4} \right)\left( {{x^2} - 4} \right)}}{{\left( {x + 2} \right)\left( {x + 4} \right)}} = \mathop {\lim }\limits_{x \to - 2} \frac{{\left( {{x^2} + 4} \right)\left( {x - 2} \right)}}{{x + 4}} = - 16\)
Câu d:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to {1^ - }} \frac{{\sqrt {1 - x} + x - 1}}{{\sqrt {{x^2} - {x^3}} }} = \mathop {\lim }\limits_{x \to {1^ - }} \frac{{\sqrt {1 - x} - \left( {1 - x} \right)}}{{\left| x \right|\sqrt {1 - x} }}\\
= \mathop {\lim }\limits_{x \to {1^ - }} \frac{{1 - \sqrt {1 - x} }}{{\left| x \right|}} = 1
\end{array}\)
Bài 32 trang 159 SGK Toán 11 nâng cao
Tìm các giới hạn sau:
a) \(\mathop {\lim }\limits_{x \to + \infty } \sqrt[3]{{\frac{{2{x^5} + {x^3} - 1}}{{\left( {2{x^2} - 1} \right)\left( {{x^3} + x} \right)}}}}\)
b) \(\mathop {\lim }\limits_{x \to - \infty } \frac{{2\left| x \right| + 3}}{{\sqrt {{x^2} + x + 5} }}\)
c) \(\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {{x^2} + x} + 2x}}{{2x + 3}}\)
d) \(\mathop {\lim }\limits_{x \to + \infty } \left( {x + 1} \right)\sqrt {\frac{x}{{2{x^4} + {x^2} + 1}}} \)
Hướng dẫn giải:
Câu a:
\(\mathop {\lim }\limits_{x \to + \infty } \sqrt[3]{{\frac{{2{x^5} + {x^3} - 1}}{{\left( {2{x^2} - 1} \right)\left( {{x^3} + x} \right)}}}} = \sqrt[3]{{\frac{{2 + \frac{1}{{{x^2}}} - \frac{1}{{{x^5}}}}}{{\left( {2 - \frac{1}{{{x^2}}}} \right)\left( {1 + \frac{1}{{{x^2}}}} \right)}}}} = 1\)
Câu b:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to - \infty } \frac{{2\left| x \right| + 3}}{{\sqrt {{x^2} + x + 5} }} = \mathop {\lim }\limits_{x \to - \infty } \frac{{2\left| x \right| + 3}}{{\left| x \right|\sqrt {1 + \frac{1}{x} + \frac{5}{{{x^2}}}} }}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{ - 2x + 3}}{{ - x\sqrt {1 + \frac{1}{x} + \frac{5}{{{x^2}}}} }} = \mathop {\lim }\limits_{x \to - \infty } \frac{{2 - \frac{3}{x}}}{{\sqrt {1 + \frac{1}{x} + \frac{5}{{{x^2}}}} }} = 2
\end{array}\)
Câu c:
\({x^2} + x \ge 0 \Leftrightarrow x \le - 1 \vee x \ge 0\)
\(\frac{{\sqrt {{x^2} + x} + 2x}}{{2x + 3}} = \frac{{\left| x \right|\sqrt {1 + \frac{1}{x}} + 2x}}{{2x + 3}} = \frac{{ - \sqrt {1 + \frac{1}{x}} + 2x}}{{2 + \frac{3}{x}}}\)
Với mọi \(x \le - 1,x \ne - \frac{3}{2}\)
Do đó \(\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {{x^2} + x} + 2x}}{{2x + 3}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - \sqrt {1 + \frac{1}{x}} + 2}}{{2 + \frac{3}{x}}} = \frac{1}{2}\)
Câu d:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to + \infty } \left( {x + 1} \right)\sqrt {\frac{x}{{2{x^4} + {x^2} + 1}}} = \mathop {\lim }\limits_{x \to + \infty } \sqrt {\frac{{x{{\left( {x + 1} \right)}^2}}}{{2{x^4} + {x^2} + 1}}} \\
= \mathop {\lim }\limits_{x \to + \infty } \sqrt {\frac{{\frac{1}{x} + \frac{2}{{{x^2}}} + \frac{1}{{{x^3}}}}}{{2 + \frac{1}{{{x^2}}} + \frac{1}{{{x^4}}}}}} = 0
\end{array}\)
Bài 33 trang 159 SGK Toán 11 nâng cao
Cho hàm số
\(f\left( x \right) = \left\{ \begin{array}{l}
{x^2} - 2x + 3,\,\,\,x \le 2\\
4x - 3,\,\,\,\,\,\,\,\,\,\,\,\,\,x > 2
\end{array} \right.\)
Tìm \(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right),\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right)\) và \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\) (nếu có)
Hướng dẫn giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \left( {4x - 3} \right) = 4.2 - 3 = 5\\
\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} \left( {{x^2} - 2x + 3} \right) = {2^2} - 2.2 + 3 = 3
\end{array}\)
Vì \(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right)\) nên không tồn tại \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\).
Trên đây là nội dung chi tiết Giải bài tập nâng cao Toán 11 Chương 4 Luyện tập (trang 159) 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.
