Giải bài 20 trang 76 sách bài tập toán 11 - Cánh diều — Không quảng cáo

Đề bài

Tính các giới hạn sau:

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

b) \(\mathop {\lim }\limits_{x \to  - 1} \frac{{ - 4x + 1}}{{{x^2} - x + 3}}\)

c) \(\mathop {\lim }\limits_{x \to 2} \sqrt {3{x^2} + 5x + 4} \)

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

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

g) \(\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} \frac{5}{{x + 2}}\)

Lời giải chi tiết

a) Ta có \(\mathop {\lim }\limits_{x \to  - 1} \left( { - 4{x^2} + 3x + 1} \right) = \mathop {\lim }\limits_{x \to  - 1} \left( { - 4{x^2}} \right) + \mathop {\lim }\limits_{x \to  - 1} 3x + \mathop {\lim }\limits_{x \to  - 1} 1 =  - 4 + \left( { - 3} \right) + 1 =  - 6\)

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

c) Xét \(\mathop {\lim }\limits_{x \to 2} \left( {3{x^2} + 5x + 4} \right) = \mathop {\lim }\limits_{x \to 2} 3{x^2} + \mathop {\lim }\limits_{x \to 2} 5x + \mathop {\lim }\limits_{x \to 2} 4 = {3.2^2} + 5.2 + 4 = 26\)

Suy ra \(\mathop {\lim }\limits_{x \to 2} \sqrt {3{x^2} + 5x + 4}  = \sqrt {26} \).

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

\( = \frac{{\mathop {\lim }\limits_{x \to  - \infty } \frac{{ - 3}}{{{x^2}}} + \mathop {\lim }\limits_{x \to  - \infty } \frac{4}{{{x^3}}}}}{{\mathop {\lim }\limits_{x \to  - \infty } 2 + \mathop {\lim }\limits_{x \to  - \infty } \frac{3}{{{x^2}}}}} = \frac{{0 + 0}}{{2 + 0}} = 0\)

e) Ta có \(\mathop {\lim }\limits_{x \to {2^ + }} \frac{{ - 3}}{{x - 2}} =  - \infty \)

f) Ta có \(\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} \frac{5}{{x + 2}} =  - \infty \)