Giải Toán 11 SGK nâng cao Chương 4 Luyện tập (trang 159)

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:

\(\mathop {\lim }\limits_{x \to  - 2} \frac{{2\left| {x + 1} \right| - 5\sqrt {{x^2} - 3} }}{{2x + 3}} = \frac{{2 - 5}}{{ - 4 + 3}} = 3\)

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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:

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

---

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:

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to  + \infty } \left( {x + 1} \right)\sqrt {\frac{x}{{2{x^4} + {x^2} + 1}}}  = \mathop {\lim }\limits_{x \to  + \infty } \sqrt {\frac{{x{{\left( {x + 1} \right)}^2}}}{{2{x^4} + {x^2} + 1}}} \\
 = \mathop {\lim }\limits_{x \to  + \infty } \sqrt {\frac{{\frac{1}{x} + \frac{2}{{{x^2}}} + \frac{1}{{{x^3}}}}}{{2 + \frac{1}{{{x^2}}} + \frac{1}{{{x^4}}}}}}  = 0
\end{array}\)

---

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 \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\).

 

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