Câu hỏi 5.12 - Mục Bài tập trang 83

1. Nội dung câu hỏi

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

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