Bài 3 trang 72 SGK Toán 11 tập 1 – Cánh Diều

Đề bài

Tính các giới hạn sau:

a) \(\mathop {\lim }\limits_{x \to 2} \left( {{x^2} – 4x + 3} \right);\)                  

b) \(\mathop {\lim }\limits_{x \to 3} \frac{{{x^2} – 5x + 6}}{{x – 3}};\)             

c) \(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x  – 1}}{{x – 1}}.\)

Lời giải chi tiết

a) \(\mathop {\lim }\limits_{x \to 2} \left( {{x^2} – 4x + 3} \right) = \mathop {\lim }\limits_{x \to 2} {x^2} – \mathop {\lim }\limits_{x \to 2} \left( {4x} \right) + 3 = {2^2} – 4.2 + 3 =  – 1\)

b) \(\mathop {\lim }\limits_{x \to 3} \frac{{{x^2} – 5x + 6}}{{x – 3}} = \mathop {\lim }\limits_{x \to 3} \frac{{\left( {x – 3} \right)\left( {x – 2} \right)}}{{x – 3}} = \mathop {\lim }\limits_{x \to 3} \left( {x – 2} \right) = \mathop {\lim }\limits_{x \to 3} x – 2 = 3 – 2 = 1\)

c) \(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x  – 1}}{{x – 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x  – 1}}{{\left( {\sqrt x  – 1} \right)\left( {\sqrt x  + 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{1}{{\sqrt x  + 1}} = \frac{1}{{\sqrt 1  + 1}} = \frac{1}{2}\)