ข้อสอบ PAT 1 - มีนาคม 2559

$\mathrm{จากสูตร} a^3-b^3 = \left(a-b\right)\left(a^2+ab+b^2\right) \to \boxed{1}$

$$\begin{gathered} \mathrm{จากโจทย์} \mathrm{ค่าของ} \underset{x\to 2^{-}}{\lim }\frac{\left|x^2-x-2\right|}{2-\sqrt[3]{x^2+4}} \\ \mathbf{พิจารณา} \left|x^2-x-2\right| = \left|\left(x-2\right)\left(x+1\right)\right| \\ = \left|x-2\right|\left|x+1\right| \end{gathered}$$

$$\begin{gathered} \mathrm{โดย} \left|x-2\right| = \left\{\begin{array}{ll}-\left(x-2\right);& \mathrm{โดย} x-2<0 \to x<2\\ x-2;& \mathrm{โดย} x-2\ge 0 \to x\ge 2\end{array}\right. \\ \left|x+1\right| = \left\{\begin{array}{ll}-\left(x+1\right);& \mathrm{โดย} x+1<0 \to x<-1\\ x+1;& \mathrm{โดย} x+1\ge 0 \to x\ge -1\end{array}\right. \\ \mathrm{แสดงว่า} x \to 2^{-} \mathrm{จะได้} \left|x-2\right|\left|x+1\right| = -\left(x-2\right)\left(x+1\right) \end{gathered}$$​

$$\begin{gathered} \mathrm{จะได้} \underset{x\to 2^{-}}{\lim }\frac{\left|x^2-x-2\right|}{2-\sqrt[3]{x^2+4}} =\underset{x\to 2^{-}}{\lim }\frac{-\left(x-2\right)\left(x+1\right)}{2-\sqrt[3]{x^2+4}} \\ = \underset{x\to 2^{-}}{\lim }\frac{-\left(x-2\right)\left(x+1\right)}{2-\sqrt[3]{x^2+4}}\times \frac{2^2+2\sqrt[3]{x^2+4}+{\left(\sqrt[3]{x^2+4}\right)}^2}{2^2+2\sqrt[3]{x^2+4}+{\left(\sqrt[3]{x^2+4}\right)}^2} \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \boxed{1} \mathrm{จะได้} \\ = \underset{x\to 2^{-}}{\lim }\frac{-\left(x-2\right)\left(x+1\right)\left(2^2+2\sqrt[3]{x^2+4}+{\left(\sqrt[3]{x^2+4}\right)}^2\right)}{2^3-{\left(\sqrt[3]{x^2+4}\right)}^3} \\ = \underset{x\to 2^{-}}{\lim }\frac{-\left(x-2\right)\left(x+1\right)\left(2^2+2\sqrt[3]{x^2+4}+{\left(\sqrt[3]{x^2+4}\right)}^2\right)}{8-\left(x^2+4\right)} \\ = \underset{x\to 2^{-}}{\lim }\frac{-\left(x-2\right)\left(x+1\right)\left(2^2+2\sqrt[3]{x^2+4}+{\left(\sqrt[3]{x^2+4}\right)}^2\right)}{8-x^2-4} \\ = \underset{x\to 2^{-}}{\lim }\frac{\left(2-x\right)\left(x+1\right)\left(2^2+2\sqrt[3]{x^2+4}+{\left(\sqrt[3]{x^2+4}\right)}^2\right)}{4-x^2} \\ = \underset{x\to 2^{-}}{\lim }\frac{\left(2-x\right)\left(x+1\right)\left(2^2+2\sqrt[3]{x^2+4}+{\left(\sqrt[3]{x^2+4}\right)}^2\right)}{\left(2-x\right)\left(2+x\right)} \end{gathered}$$

$$\begin{gathered} = \underset{x\to 2^{-}}{\lim }\frac{\left(x+1\right)\left(2^2+2\sqrt[3]{x^2+4}+{\left(\sqrt[3]{x^2+4}\right)}^2\right)}{2+x} \\ = \frac{\left(2+1\right)\left(2^2+2\sqrt[3]{2^2+4}+{\left(\sqrt[3]{2^2+4}\right)}^2\right)}{2+2} \\ = \frac{\left(3\right)[4+2\left(2\right)+{\left(2\right)}^2]}{4} \\ = 9 \\ \mathrm{ค่าของ} \underset{x\to 2^{-}}{lim}\frac{\left|x^2-x-2\right|}{2-\sqrt[3]{x^2+4}} = 9 \end{gathered}$$