Bài 5.11 trang 118 SGK Toán 11 tập 1 – Kết nối tri thức

Đề bài

Cho hàm số \(g\left( x \right) = \frac{{{x^2} – 5x + 6}}{{\left| {x – 2} \right|}}\)

Tìm \(\mathop {{\rm{lim}}}\limits_{x \to {2^ + }} g\left( x \right)\) và \(\mathop {{\rm{lim}}}\limits_{x \to {2^ – }} g\left( x \right)\).

Lời giải chi tiết

Khi \(x \to {2^ – } \Rightarrow \left| {x – 2} \right| = 2 – x\)

Ta có:

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

Khi \(x \to {2^ + } \Rightarrow \left| {x – 2} \right| = x – 2\)

Ta có

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