ข้อสอบ PAT 1 - กุมภาพันธ์ 2562

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

$$\begin{gathered} \mathrm{จากโจทย์} \underset{x\to 0}{\lim }\frac{x\sqrt{x+\sqrt{1+x}}}{\sqrt[3]{8+0}-2} \to \boxed{2} \\ = \frac{0\sqrt{0+\sqrt{1+0}}}{\sqrt[3]{8+0}-2} \\ = \frac{0}{0} \\ \mathrm{แสดงว่า} \mathrm{ต้องจัดรูปลิมิตใหม่} \end{gathered}$$

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

$$\begin{gathered} = \frac{x\sqrt{x+\sqrt{1+x}}\left[{\left(\sqrt[3]{8+x}\right)}^2+\sqrt[3]{8+x}\left(2\right)+4\right]}{x} \\ = x\sqrt{x+\sqrt{1+x}}\left[{\left(\sqrt[3]{8+x}\right)}^2+\sqrt[3]{8+x}\left(2\right)+4\right] \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \boxed{2} \mathrm{จะได้} \\ \underset{x\to 0}{\lim }\frac{x\sqrt{x\sqrt{1+x}}}{\sqrt[3]{8+x}-2} \\ = \underset{x\to 0}{\lim }x\sqrt{x\sqrt{1+x}}\left[{\left(\sqrt[3]{8+x}\right)}^2+\sqrt[3]{8+x}\left(2\right)+4\right] \end{gathered}$$

$$\begin{gathered} = \sqrt{0+\sqrt{1+0}}\left[{\left(\sqrt[3]{8+0}\right)}^2+\sqrt[3]{8+0}\left(2\right)+4\right] \\ = 1\left(2^2+2\left(2\right)+4\right) \\ = 12 \\ \mathrm{ดังนั้น} \underset{x\to 0}{\lim }\frac{x\sqrt{x+\sqrt{1+x}}}{\sqrt[3]{8+x}-2} = 12 \end{gathered}$$