Bài 5.28 trang 124 SGK Toán 11 tập 1 – Kết nối tri thức

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) \(\mathop {\lim }\limits_{x \to  – \infty } \frac{{x + 2}}{{\sqrt {4{x^2} + 1} }} = \mathop {\lim }\limits_{x \to  – \infty } \frac{{1 + \frac{2}{x}}}{{\sqrt {4 + \frac{1}{{{x^2}}}} }} =  – \frac{1}{2}\)