Bài tập 4.23 trang 165 SBT Toán 11

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

b) \(\mathop {\lim }\limits_{x \to  + \infty } \frac{{x - 1}}{{{x^2} - 1}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{x - 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{1}{{x + 1}} = 0\)

c) \(\mathop {\lim }\limits_{x \to 5} \frac{{x - 5}}{{\sqrt x  - \sqrt 5 }} = \mathop {\lim }\limits_{x \to 5} \frac{{\left( {\sqrt x  - \sqrt 5 } \right)\left( {\sqrt x  + \sqrt 5 } \right)}}{{\sqrt x  - \sqrt 5 }} = \mathop {\lim }\limits_{x \to 5} \left( {\sqrt x  + \sqrt 5 } \right) = 2\sqrt 5 \)

d) \(\mathop {\lim }\limits_{x \to  + \infty } \frac{{x - 5}}{{\sqrt x  + \sqrt 5 }} = \mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt x  - \sqrt 5 } \right) =  + \infty \)

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

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

g) \(\mathop {\lim }\limits_{x \to 0} \frac{1}{{{x^2}}}\left( {\frac{1}{{{x^2} + 1}} - 1} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{{{x^2}}}.\frac{{ - {x^2}}}{{{x^2} + 1}}} \right) = \mathop {\lim }\limits_{x \to 0} \left( { - \frac{1}{{{x^2} + 1}}} \right) =  - 1\)

h) 

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to  - \infty } \frac{{({x^2} - 1){{(1 - 2x)}^5}}}{{{x^7} + x + 3}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{{x^2}(1 - \frac{1}{{{x^2}}}).{x^5}{{(\frac{1}{x} - 2)}^5}}}{{{x^7}(1 + \frac{1}{{{x^6}}} + \frac{3}{{{x^7}}})}}\\
 = \mathop {\lim }\limits_{x \to  - \infty } \frac{{(1 - \frac{1}{{{x^2}}}){{(\frac{1}{x} - 2)}^5}}}{{1 + \frac{1}{{{x^6}}} + \frac{3}{{{x^7}}}}} = {( - 2)^5} =  - 32
\end{array}\)

-- Mod Toán 11 HỌC247