Bài 5. 28 trang 124 SGK Toán 11 tập 1 - Kết nối tri thức — Không quảng cáo

Đề bài

Tính các giới hạn sau:

a) $\mathop {{\rm{lim}}}\limits_{x \to 7} \frac{{\sqrt {x + 2}  - 3}}{{x - 7}}$;

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

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

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

Lời giải chi tiết

a) $\mathop {\lim }\limits_{x \to 7} \frac{{\sqrt {x + 2}  - 3}}{{x - 7}} = \mathop {\lim }\limits_{x \to 7} \frac{1}{{\sqrt {x + 2}  + 3}} = \frac{1}{6}$

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

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

$\mathop {\lim }\limits_{x \to 1} \left( {2 - x} \right) = 1$

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

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

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