ข้อสอบคณิต 9 สามัญ - ปี 2562

$$\begin{gathered} \mathrm{จากโจทย์} \underset{x\to 2^{-}}{\lim }\frac{\left|x-2\right|}{x^2+5x-14} \to \boxed{1} \\ - \mathrm{แทนค่า} x=2 \\ \mathrm{จะได้} \underset{x\to 2^{-}}{\lim }\frac{\left|x-2\right|}{x^2+5x-14} = \frac{\left|2-2\right|}{2^2+5\left(2\right)-14} \\ = \frac{0}{0} \\ \mathrm{แสดงว่า} \mathrm{ต้องจัดรูป} limit \mathrm{ใหม่} \end{gathered}$$

$$\begin{gathered} \mathrm{นิยาม} \\ \left|x-2\right| = \left\{\begin{array}{ll}x-2& \mathrm{เมื่อ} x-2 \ge 0 \to x\ge 2\\ -\left(x-2\right)& \mathrm{เมื่อ} x-2 < 0 \to x<2\end{array}\right. \\ - \mathrm{โดย} x\to 2^{-} \mathrm{คือ} x\approx 1,999 \left(x<2\right) ; \mathrm{แทนใน} \boxed{1} \\ \mathrm{จะได้} \left|x-2\right| = -\left(x-2\right) \end{gathered}$$

$$\begin{gathered} \underset{x\to 2^{-}}{\lim }\frac{\left|x-2\right|}{x^2+5x-14} = \underset{x\to 2^{-}}{\lim }\frac{-\left(x-2\right)}{x^2+5x-14} \\ = \underset{x\to 2^{-}}{\lim }\frac{-\cancel{\left(x-2\right)}}{\cancel{\left(x-2\right)}\left(x+7\right)} \\ = \underset{x\to 2^{-}}{\lim }\frac{-1}{x+7} ; \mathrm{แทน} x=2 \\ = \frac{-1}{2+7} = \frac{-1}{9} \end{gathered}$$