Giải bài 3 trang 84 sách bài tập toán 11 – Chân trời sáng tạo tập 1

Đề bài

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

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