Giải bài 5.12 trang 83 sách bài tập toán 11 – Kết nối tri thức với cuộc sống

Đề bài

Tính các giới hạn sau:

a) \(\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {4x + 1}  – 3}}{{x – 2}};\)                         

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

c) \(\mathop {\lim }\limits_{x \to {2^ + }} \frac{{{x^2} – 5x + 6}}{{{{\left( {x – 2} \right)}^2}}};\)              

d) \(\mathop {\lim }\limits_{x \to {0^ – }} \frac{{{x^2} + x – 2}}{x}.\)

Lời giải chi tiết

a) \(\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {4x + 1}  – 3}}{{x – 2}} = \mathop {\lim }\limits_{x \to 2} \frac{{4x – 8}}{{\left( {x – 2} \right)\left( {\sqrt {4x + 1}  + 3} \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{4}{{\sqrt {4x + 1}  + 3}} = \frac{4}{{\sqrt {4.2 + 1}  + 3}} = \frac{2}{3}\)       

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

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

c) \(\mathop {\lim }\limits_{x \to {2^ + }} \frac{{{x^2} – 5x + 6}}{{{{\left( {x – 2} \right)}^2}}} = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{\left( {x – 2} \right)\left( {x – 3} \right)}}{{{{\left( {x – 2} \right)}^2}}} = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{x – 3}}{{x – 2}}\)      

Vì \(\mathop {\lim }\limits_{x \to {2^ + }} \left( {x – 3} \right) =  – 1,\mathop {\lim }\limits_{x \to {2^ + }} \left( {x – 2} \right) = 0\) và \(x – 2 > 0\;\forall x > 2\) nên \(\mathop {\lim }\limits_{x \to {2^ + }} \frac{{x – 3}}{{x – 2}} =  – \infty \)

d) \(\mathop {\lim }\limits_{x \to {0^ – }} \frac{{{x^2} + x – 2}}{x}\)

Vì \(\mathop {\lim }\limits_{x \to {0^ – }} \left( {{x^2} + x – 2} \right) =  – 2,\mathop {\lim }\limits_{x \to {0^ – }} x = 0\) và \(x < 0\) \(\forall x < 0\) nên  \(\mathop {\lim }\limits_{x \to {0^ – }} \frac{{{x^2} + x – 2}}{x} =  + \infty \)