Tính các giới hạn sau: a) lim x đến 7 căn bậc hai của x + 2  - 3/x - 7; b) lim x đến 1x^3 - 1/x^2 - 1; c) lim x đến 12 - x/ 1 - x

Tính các giới hạn sau:

a) lim;

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

c) \mathop {\lim }\limits_{x \to 1} \frac{{2 - x}}{{{{\left( {1 - x} \right)}^2}}};

d) \mathop {\lim }\limits_{x \to - \infty } \frac{{x + 2}}{{\sqrt {4{x^2} + 1} }}.

Trả lời

Lời giải:

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

= \mathop {\lim }\limits_{x \to 7} \frac{{x - 7}}{{\left( {x - 7} \right)\left( {\sqrt {x + 2} + 3} \right)}} = \mathop {\lim }\limits_{x \to 7} \frac{1}{{\sqrt {x + 2} + 3}} = \frac{1}{{\sqrt {7 + 2} + 3}} = \frac{1}{6}.

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

c) \mathop {\lim }\limits_{x \to 1} \frac{{2 - x}}{{{{\left( {1 - x} \right)}^2}}}

Ta có: \mathop {\lim }\limits_{x \to 1} \left( {2 - x} \right) = 2 - 1 = 1 > 0;

\mathop {\lim }\limits_{x \to 1} {\left( {1 - x} \right)^2} = 0 và (1 – x)2 > 0 với mọi x ≠ 1.

Do vậy, \mathop {\lim }\limits_{x \to 1} \frac{{2 - x}}{{{{\left( {1 - x} \right)}^2}}} = + \infty .

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

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

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