Câu hỏi 3 - Mục Bài tập trang 84

1. Nội dung câu hỏi

Tìm các giới hạn sau:

a) \(\mathop {\lim }\limits_{x \to  - 2} \frac{{{x^2} - 4}}{{x + 2}}\);

b) \(\mathop {\lim }\limits_{x \to 1} \frac{{{x^3} - 1}}{{1 - x}}\);

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

d) \(\mathop {\lim }\limits_{x \to  - 2} \frac{{2 - \sqrt {x + 6} }}{{x + 2}}\);

e) \(\mathop {\lim }\limits_{x \to 0} \frac{x}{{\sqrt {x + 1}  - 1}}\);

g) \(\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 4x + 4}}{{{x^2} - 4}}\).

3. Lời giải chi tiết 

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

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

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

d) \(\mathop {\lim }\limits_{x \to  - 2} \frac{{2 - \sqrt {x + 6} }}{{x + 2}} \) \( = \mathop {\lim }\limits_{x \to  - 2} \frac{{\left( {2 - \sqrt {x + 6} } \right)\left( {2 + \sqrt {x + 6} } \right)}}{{\left( {x + 2} \right)\left( {2 + \sqrt {x + 6} } \right)}} \) \( = \mathop {\lim }\limits_{x \to  - 2} \frac{{4 - x - 6}}{{\left( {x + 2} \right)\left( {2 + \sqrt {x + 6} } \right)}}\)

\( = \mathop {\lim }\limits_{x \to  - 2} \frac{{ - \left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {2 + \sqrt {x + 6} } \right)}} \) \( = \mathop {\lim }\limits_{x \to  - 2} \frac{{ - 1}}{{2 + \sqrt {x + 6} }} \) \( = \frac{{ - 1}}{{2 + \sqrt { - 2 + 6} }} \) \( = \frac{{ - 1}}{4}\)

e) \(\mathop {\lim }\limits_{x \to 0} \frac{x}{{\sqrt {x + 1}  - 1}} \) \( = \mathop {\lim }\limits_{x \to 0} \frac{{x\left( {\sqrt {x + 1}  + 1} \right)}}{{\left( {\sqrt {x + 1}  - 1} \right)\left( {\sqrt {x + 1}  + 1} \right)}} \) \( = \mathop {\lim }\limits_{x \to 0} \frac{{x\left( {\sqrt {x + 1}  + 1} \right)}}{x}\)

\( = \mathop {\lim }\limits_{x \to 0} \left( {\sqrt {x + 1}  + 1} \right) \) \( = \sqrt {0 + 1}  + 1 \) \( = 2\);

g) \(\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 4x + 4}}{{{x^2} - 4}} \) \( = \mathop {\lim }\limits_{x \to 2} \frac{{{{\left( {x - 2} \right)}^2}}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} \) \( = \mathop {\lim }\limits_{x \to 2} \frac{{x - 2}}{{x + 2}} \) \( = \frac{{2 - 2}}{{2 + 2}} \) \( = 0\).