Bài 17 trang 234 SBT đại số và giải tích 11

Tính các giới hạn sau

LG a

\(\mathop {\lim }\limits_{x \to 1} \frac{{4{x^5} + 9x + 7}}{{3{x^6} + {x^3} + 1}}\)

Lời giải chi tiết:

\(\mathop {\lim }\limits_{x \to 1} \frac{{4{x^5} + 9x + 7}}{{3{x^6} + {x^3} + 1}}\) \( = \frac{{{{4.1}^5} + 9.1 + 7}}{{{{3.1}^6} + {1^3} + 1}} = 4\)

LG b

\(\mathop {\lim }\limits_{x \to 2} \frac{{{x^3} + 3{x^2} - 9x - 2}}{{{x^3} - x - 6}}\)

Lời giải chi tiết:

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 2} \frac{{{x^3} + 3{x^2} - 9x - 2}}{{{x^3} - x - 6}}\\ = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x - 2} \right)\left( {{x^2} + 5x + 1} \right)}}{{\left( {x - 2} \right)\left( {{x^2} + 2x + 3} \right)}}\\ = \mathop {\lim }\limits_{x \to 2} \frac{{{x^2} + 5x + 1}}{{{x^2} + 2x + 3}}\\ = \frac{{{2^2} + 5.2 + 1}}{{{2^2} + 2.2 + 3}} = \frac{{15}}{{11}}\end{array}\)

LG c

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

Lời giải chi tiết:

\(\begin{array}{l}\mathop {\lim }\limits_{x \to  - 1} \frac{{x + 1}}{{\sqrt {6{x^2} + 3}  + 3x}}\\ = \mathop {\lim }\limits_{x \to  - 1} \frac{{\left( {x + 1} \right)\left( {\sqrt {6{x^2} + 3}  - 3x} \right)}}{{6{x^2} + 3 - 9{x^2}}}\\ = \mathop {\lim }\limits_{x \to  - 1} \frac{{\left( {x + 1} \right)\left( {\sqrt {6{x^2} + 3}  - 3x} \right)}}{{3 - 3{x^2}}}\\ = \mathop {\lim }\limits_{x \to  - 1} \frac{{\left( {x + 1} \right)\left( {\sqrt {6{x^2} + 3}  - 3x} \right)}}{{3\left( {1 - x} \right)\left( {1 + x} \right)}}\\ = \mathop {\lim }\limits_{x \to  - 1} \frac{{\left( {\sqrt {6{x^2} + 3}  - 3x} \right)}}{{3\left( {1 - x} \right)}}\\ = \frac{{\sqrt {6.{{\left( { - 1} \right)}^2} + 3}  - 3.\left( { - 1} \right)}}{{3\left( {1 + 1} \right)}}\\ = 1\end{array}\)

LG d

\(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {9 + 5x + 4{x^2}}  - 3}}{x}\)

Lời giải chi tiết:

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {9 + 5x + 4{x^2}}  - 3}}{x}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{9 + 5x + 4{x^2} - 9}}{{x\left( {\sqrt {9 + 5x + 4{x^2}}  + 3} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{5x + 4{x^2}}}{{x\left( {\sqrt {9 + 5x + 4{x^2}}  + 3} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{x\left( {5 + 4x} \right)}}{{x\left( {\sqrt {9 + 5x + 4{x^2}}  + 3} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {5 + 4x} \right)}}{{\left( {\sqrt {9 + 5x + 4{x^2}}  + 3} \right)}}\\ = \frac{5}{6}\end{array}\)

LG e

\(\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt[3]{{10 - x}} - 2}}{{x - 2}}\)

Lời giải chi tiết:

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt[3]{{10 - x}} - 2}}{{x - 2}}\\ = \mathop {\lim }\limits_{x \to 2} \frac{{10 - x - 8}}{{\left( {x - 2} \right)\left[ {{{\left( {\sqrt[3]{{10 - x}}} \right)}^2} + 2\sqrt[3]{{10 - x}} + 4} \right]}}\\ = \mathop {\lim }\limits_{x \to 2} \frac{{2 - x}}{{\left( {x - 2} \right)\left[ {{{\left( {\sqrt[3]{{10 - x}}} \right)}^2} + 2\sqrt[3]{{10 - x}} + 4} \right]}}\\ = \mathop {\lim }\limits_{x \to 2} \frac{{ - \left( {x - 2} \right)}}{{\left( {x - 2} \right)\left[ {{{\left( {\sqrt[3]{{10 - x}}} \right)}^2} + 2\sqrt[3]{{10 - x}} + 4} \right]}}\\ = \mathop {\lim }\limits_{x \to 2} \frac{{ - 1}}{{{{\left( {\sqrt[3]{{10 - x}}} \right)}^2} + 2\sqrt[3]{{10 - x}} + 4}}\\ =  - \frac{1}{{12}}\end{array}\)

LG f

\(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {x + 8}  - \sqrt {8x + 1} }}{{\sqrt {5 - x}  - \sqrt {7x - 3} }}\)

Lời giải chi tiết:

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {x + 8}  - \sqrt {8x + 1} }}{{\sqrt {5 - x}  - \sqrt {7x - 3} }}\\ = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x + 8 - 8x - 1} \right)\left( {\sqrt {5 - x}  + \sqrt {7x - 3} } \right)}}{{\left( {5 - x - 7x + 3} \right)\left( {\sqrt {x + 8}  + \sqrt {8x + 1} } \right)}}\\ = \mathop {\lim }\limits_{x \to 1} \frac{{\left( { - 7x + 7} \right)\left( {\sqrt {5 - x}  + \sqrt {7x - 3} } \right)}}{{\left( { - 8x + 8} \right)\left( {\sqrt {x + 8}  + \sqrt {8x + 1} } \right)}}\\ = \mathop {\lim }\limits_{x \to 1} \frac{{ - 7\left( {x - 1} \right)\left( {\sqrt {5 - x}  + \sqrt {7x - 3} } \right)}}{{ - 8\left( {x - 1} \right)\left( {\sqrt {x + 8}  + \sqrt {8x + 1} } \right)}}\\ = \mathop {\lim }\limits_{x \to 1} \frac{{7\left( {\sqrt {5 - x}  + \sqrt {7x - 3} } \right)}}{{8\left( {\sqrt {x + 8}  + \sqrt {8x + 1} } \right)}}\\ = \frac{7}{{12}}\end{array}\)