ข้อสอบ PAT 1 - กุมภาพันธ์ 2563

$$\begin{gathered} \mathrm{แทน} x=1 \mathrm{จะได้} \frac{0}{0} \\ \mathrm{ต้องจัดรูปให้} \left(x-1\right) \mathrm{ตัดกัน} \\ \mathrm{ตัวเศษ} \mathrm{ต้องคูณ} \sqrt{x}+1 \mathrm{และตัวส่วนต้องแยกตัวประกอบ} \end{gathered}$$

$$\begin{gathered} \underset{x\to 1}{\lim }\frac{\left(\sqrt{x}-1\right)\left(3x-2\right)}{3x^2-x-2} = \underset{x\to 1}{\lim }\frac{\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)}·\frac{\left(\sqrt{x}-1\right)\left(3x-2\right)}{\left(x-1\right)\left(3x+2\right)} \\ = \underset{x\to 1}{\lim }\frac{\cancel{\left(x-1\right)}\left(3x-2\right)}{\left(\sqrt{x}+1\right)\cancel{\left(x-1\right)}\left(3x+2\right)} \end{gathered}$$

$$\begin{gathered} = \underset{x\to 1}{\lim }\frac{\left(3x-2\right)}{\left(\sqrt{x}+1\right)\left(3x+2\right)} \\ = \frac{3\left(1\right)-2}{\left(\sqrt{1}+1\right)\left(3\left(1\right)+2\right)} = \frac{1}{10} \to \mathrm{ตอบ} 3 \end{gathered}$$