Đề bài

Tìm các giới hạn: a) \(\mathop {\lim }\limits_{x \to 4} \frac{{{x^2} - x - 12}}{{{x^2} - 16}}\) b) \(\mathop {\lim }\limits_{x \to 1} \frac{{{x^4} - 1}}{{{x^3} - 1}}\) c) \(\mathop {\lim }\limits_{x \to  + \infty } \frac{{{x^3} + x + 5}}{{2{x^3} - 1}}\) d) \(\mathop {\lim }\limits_{x \to  - \infty } \frac{{\sqrt {1 + {x^2}} }}{{2x - 1}}\)

Lời giải chi tiết

a, 

\(\mathop {\lim }\limits_{x \to 4} \frac{{{x^2} - x - 12}}{{{x^2} - 16}} = \mathop {\lim }\limits_{x \to 4} \frac{{\left( {x + 3} \right)\left( {x - 4} \right)}}{{\left( {x + 4} \right)\left( {x - 4} \right)}} = \mathop {\lim }\limits_{x \to 4} \frac{{x + 3}}{{x + 4}} = \frac{{4 + 3}}{{4 + 4}} = \frac{7}{8}\)

b, 

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

c, 

Chia cả từ và mẫu cho \({x^3}\) ta được \(\mathop {\lim }\limits_{x \to  + \infty } \frac{{{x^3} + x + 5}}{{2{x^3} - 1}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{1 + \frac{1}{{{x^2}}} + \frac{5}{{{x^3}}}}}{{2 - \frac{1}{{{x^3}}}}} = \frac{1}{2}\)

d, 

Chia cả tử và mẫu cho \(x\)\(\mathop {\lim }\limits_{x \to  - \infty } \frac{{\sqrt {1 + {x^2}} }}{{2x - 1}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{\left| x \right|\sqrt {\frac{1}{{{x^2}}} + 1} }}{{2x - 1}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{ - x\sqrt {\frac{1}{{{x^2}}} + 1} }}{{2x - 1}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{ - \sqrt {\frac{1}{{{x^2}}} + 1} }}{{2 - \frac{1}{x}}} =  - \frac{1}{2}\)