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Câu hỏi: 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 \(\mathop {{\rm{lim}}}\limits_{x \to {2^ - }} g\left( x \right)\).
Tìm \(\mathop {{\rm{lim}}}\limits_{x \to {2^ + }} g\left( x \right)\) và \(\mathop {{\rm{lim}}}\limits_{x \to {2^ - }} g\left( x \right)\).
Phương pháp giải
Áp dụng giới hạn trái, phải để tính.
\(\left| a \right| = \left\{ \begin{array}{l} - a,a < 0\\a,a \ge 0\end{array} \right.\)
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
\(\mathop {\lim }\limits_{x \to {2^ + }} \frac{{{x^2} - 5x + 6}}{{\left| {x - 2} \right|}} = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{{x^2} - 5x + 6}}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{\left( {x - 2} \right)\left( {x - 3} \right)}}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ - }} \left( {x - 3} \right) = 2 - 3 = - 1\)
Áp dụng giới hạn trái, phải để tính.
\(\left| a \right| = \left\{ \begin{array}{l} - a,a < 0\\a,a \ge 0\end{array} \right.\)
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
\(\mathop {\lim }\limits_{x \to {2^ + }} \frac{{{x^2} - 5x + 6}}{{\left| {x - 2} \right|}} = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{{x^2} - 5x + 6}}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{\left( {x - 2} \right)\left( {x - 3} \right)}}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ - }} \left( {x - 3} \right) = 2 - 3 = - 1\)