Bài 18 trang 234 SBT đại số và giải tích 11

Tính các giới hạn

LG a

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

Lời giải chi tiết:

\(\begin{array}{l}\mathop {\lim }\limits_{x \to  + \infty } \left( {\frac{{{x^3}}}{{3{x^2} - 4}} - \frac{{{x^2}}}{{3x + 2}}} \right)\\ = \mathop {\lim }\limits_{x \to  + \infty } \frac{{3{x^4} + 2{x^3} - 3{x^4} + 4{x^2}}}{{\left( {3{x^2} - 4} \right)\left( {3x + 2} \right)}}\\ = \mathop {\lim }\limits_{x \to  + \infty } \frac{{2{x^3} + 4{x^2}}}{{9{x^3} + 6{x^2} - 12x - 8}}\\ = \mathop {\lim }\limits_{x \to  + \infty } \frac{{{x^3}\left( {2 + \frac{4}{x}} \right)}}{{{x^3}\left( {9 + \frac{6}{x} - \frac{{12}}{x} - \frac{8}{{{x^3}}}} \right)}}\\ = \mathop {\lim }\limits_{x \to  + \infty } \frac{{2 + \frac{4}{x}}}{{9 + \frac{6}{x} - \frac{{12}}{x} - \frac{8}{{{x^3}}}}}\\ = \frac{{2 + 0}}{{9 + 0 - 0 - 8}}\\ = \frac{2}{9}\end{array}\)

LG b

\(\mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {9{x^2} + 1}  - 3x} \right)\)

Lời giải chi tiết:

\(\begin{array}{l}\mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {9{x^2} + 1}  - 3x} \right)\\ = \mathop {\lim }\limits_{x \to  + \infty } \frac{{9{x^2} + 1 - 9{x^2}}}{{\sqrt {9{x^2} + 1}  + 3x}}\\ = \mathop {\lim }\limits_{x \to  + \infty } \frac{1}{{\sqrt {9{x^2} + 1}  + 3x}}\\ = 0\end{array}\)

Vì \(\mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {9{x^2} + 1}  + 3x} \right) =  + \infty \).

LG c

\(\mathop {\lim }\limits_{x \to  - \infty } \left( {\sqrt {2{x^2} - 3}  - 5x} \right)\)

Lời giải chi tiết:

\(\begin{array}{l}\mathop {\lim }\limits_{x \to  - \infty } \left( {\sqrt {2{x^2} - 3}  - 5x} \right)\\ = \mathop {\lim }\limits_{x \to  - \infty } \left[ {\sqrt {{x^2}\left( {2 - \frac{3}{{{x^2}}}} \right)}  - 5x} \right]\\ = \mathop {\lim }\limits_{x \to  - \infty } \left[ {\left| x \right|\sqrt {2 - \frac{3}{{{x^2}}}}  - 5x} \right]\\ = \mathop {\lim }\limits_{x \to  - \infty } \left[ { - x\sqrt {2 - \frac{3}{{{x^2}}}}  - 5x} \right]\\ = \mathop {\lim }\limits_{x \to  - \infty } \left[ { - x\left( {\sqrt {2 - \frac{3}{{{x^2}}}}  + 5} \right)} \right]\\ =  + \infty \end{array}\)

Vì \(\mathop {\lim }\limits_{x \to  - \infty } \left( { - x} \right) =  + \infty \) và \(\mathop {\lim }\limits_{x \to  - \infty } \left( {\sqrt {2 - \frac{3}{{{x^2}}}}  + 5} \right) = \sqrt 2  + 5 > 0\).

LG d

\(\mathop {\lim }\limits_{x \to  + \infty } \frac{{\sqrt {2{x^2} + 3} }}{{4x + 2}}\)

Lời giải chi tiết:

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

LG e

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

Lời giải chi tiết:

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