Bài 3 trang 79 SGK Toán 11 tập 1 - Cánh Diều

Đề bài

Tính các giới hạn sau:

a) $\mathop {\lim }\limits_{x \to  - 3} \left( {4{x^2} - 5x + 6} \right)$;           

b) $\mathop {\lim }\limits_{x \to 2} \frac{{2{x^2} - 5x + 2}}{{x - 2}}$;                  

c) $\mathop {\lim }\limits_{x \to 4} \frac{{\sqrt x  - 2}}{{{x^2} - 16}}$.

Lời giải chi tiết

a) $\mathop {\lim }\limits_{x \to  - 3} \left( {4{x^2} - 5x + 6} \right) = 4.{\left( { - 3} \right)^2} - 5.\left( { - 3} \right) + 6 = 57$

b) $\mathop {\lim }\limits_{x \to 2} \frac{{2{x^2} - 5x + 2}}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x - 2} \right)\left( {2x - 1} \right)}}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \left( {2x - 1} \right) = 2.2 - 1 = 3$

c) $\begin{array}{c}\mathop {\lim }\limits_{x \to 4} \frac{{\sqrt x  - 2}}{{{x^2} - 16}} = \mathop {\lim }\limits_{x \to 4} \frac{{\sqrt x  - 2}}{{\left( {x - 4} \right)\left( {x + 4} \right)}} = \mathop {\lim }\limits_{x \to 4} \frac{{\sqrt x  - 2}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)\left( {x + 4} \right)}} = \mathop {\lim }\limits_{x \to 4} \frac{1}{{\left( {\sqrt x  + 2} \right)\left( {x + 4} \right)}}\\ = \frac{1}{{\left( {\sqrt 4  + 2} \right)\left( {4 + 4} \right)}} = \frac{1}{{32}}\end{array}$