AMBIENT
  • Câu hỏi:

    Tính các giới hạn sau:

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

    b) \(\mathop {lim}\limits_{x \to 2} \frac{{\sqrt {{x^2} + 12}  - 4}}{{{x^2} - 3x + 2}}\)

    c) \(\lim \left( {\sqrt {9{n^2} - 3n + 1}  - 3n} \right)\)

    d) \(\mathop {lim}\limits_{x \to  - \infty } \frac{{\sqrt {{x^2} + x - 2}  + 2x - 1}}{{x - 1}}\)

    e) \(\mathop {lim}\limits_{x \to 2} \frac{{\sqrt {{x^2} - 3}  - 1}}{{\sqrt {x + 7}  - 3}}\)

    Lời giải tham khảo:

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

    \( = \mathop {lim}\limits_{x \to 1} \frac{{\left( {{x^3} + {x^2} + x + 2} \right)}}{{\left( {x + 1} \right)}} = \frac{5}{2}\)

    b) \(\mathop {lim}\limits_{x \to 2} \frac{{\sqrt {{x^2} + 12}  - 4}}{{{x^2} - 3x + 2}} = \mathop {lim}\limits_{x \to 2} \frac{{{x^2} + 12 - 16}}{{\left( {x - 2} \right)\left( {x - 1} \right)\left( {\sqrt {{x^2} + 12}  + 4} \right)}}\)

    \( = \mathop {lim}\limits_{x \to 2} \frac{{{x^2} - 4}}{{\left( {x - 2} \right)\left( {x - 1} \right)\left( {\sqrt {{x^2} + 12}  + 4} \right)}}\)

    \( = \mathop {lim}\limits_{x \to 2} \frac{{\left( {x + 2} \right)\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x - 1} \right)\left( {\sqrt {{x^2} + 12}  + 4} \right)}} = \mathop {lim}\limits_{x \to 2} \frac{{\left( {x + 2} \right)}}{{\left( {x - 1} \right)\left( {\sqrt {{x^2} + 12}  + 4} \right)}} = \frac{4}{8} = \frac{1}{2}\)

    c) \(\lim \left( {\sqrt {9{n^2} - 3n + 1}  - 3n} \right) = \lim \frac{{9{n^2} - 3n + 1 - 9{n^2}}}{{\left( {\sqrt {9{n^2} - 3n + 1}  + 3n} \right)}} = \lim \frac{{ - 3n + 1}}{{\left( {\sqrt {9{n^2} - 3n + 1}  + 3n} \right)}}\)

    \( = \lim \frac{{ - 3 + \frac{1}{n}}}{{\left( {\sqrt {9 - \frac{3}{n} + \frac{1}{{{n^2}}}}  + 3} \right)}} =  - \frac{1}{2}\)

    d) \(\mathop {lim}\limits_{x \to  - \infty } \frac{{\sqrt {{x^2} + x - 2}  + 2x - 1}}{{x - 1}} = \mathop {lim}\limits_{x \to  - \infty } \frac{{ - \sqrt {1 + \frac{x}{x} - \frac{2}{{{x^2}}}}  + 2 - \frac{1}{x}}}{{1 - \frac{1}{x}}} = 1\)

    e) \(\mathop {lim}\limits_{x \to 2} \frac{{\sqrt {{x^2} - 3}  - 1}}{{\sqrt {x + 7}  - 3}} = \mathop {lim}\limits_{x \to 2} \frac{{\left( {{x^2} - 3 - 1} \right)\left( {\sqrt {x + 7}  + 3} \right)}}{{\left( {x + 7 - 9} \right)\left( {\sqrt {{x^2} - 3}  + 1} \right)}} = \mathop {lim}\limits_{x \to 2} \frac{{\left( {{x^2} - 4} \right)\left( {\sqrt {x + 7}  + 3} \right)}}{{\left( {x - 2} \right)\left( {\left( {\sqrt {{x^2} - 3}  + 1} \right)} \right)}}\)

    \( = \mathop {lim}\limits_{x \to 2} \frac{{\left( {x - 2} \right)\left( {x + 2} \right)\left( {\sqrt {x + 7}  + 3} \right)}}{{\left( {x - 2} \right)\left( {\sqrt {{x^2} - 3}  + 1} \right)}} = \mathop {lim}\limits_{x \to 2} \frac{{\left( {x + 2} \right)\left( {\sqrt {x + 7}  + 3} \right)}}{{\left( {\sqrt {{x^2} - 3}  + 1} \right)}} = 12\)

    ADSENSE

Câu hỏi này thuộc đề thi trắc nghiệm dưới đây, bấm vào Bắt đầu thi để làm toàn bài

Hướng dẫn Trắc nghiệm Online và Tích lũy điểm thưởng

 

 

CÂU HỎI KHÁC

AMBIENT
?>