ข้อสอบ PAT 1 - มีนาคม 2556

$$\begin{gathered} \mathrm{จากโจทย์}\hspace{0.17em} \underset{\mathrm{x}\to \infty }{\lim }\left(\sqrt{\mathrm{x}\left(\mathrm{x}-1\right)}-\mathrm{x}+2\right) \\ =\underset{\mathrm{x}\to \infty }{\lim }\left(\sqrt{\mathrm{x}\left(\mathrm{x}-1\right)}-\mathrm{x}\right)+\underset{\mathrm{x}\to \infty }{\lim }\left(2\right) \\ =2+\underset{\mathrm{x}\to \infty }{\lim }\left(\sqrt{\mathrm{x}\left(\mathrm{x}-1\right)}-\mathrm{x}\right) \end{gathered}$$

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

$$\begin{gathered} \mathrm{นำ} \mathrm{x} \mathrm{หารทั้งเศษและส่วน} \\ =2+\underset{\mathrm{x}\to \infty }{\lim }\frac{-1}{\sqrt{{\displaystyle \frac{\mathrm{x}\left(\mathrm{x}-1\right)}{{\mathrm{x}}^2}}}+1} \\ =2+\underset{\mathrm{x}\to \infty }{\lim }\frac{-1}{\sqrt{{\displaystyle \frac{{\mathrm{x}}^2-\mathrm{x}}{{\mathrm{x}}^2}}}+1} \end{gathered}$$

$$\begin{gathered} =2+\underset{\mathrm{x}\to \infty }{\lim }\frac{-1}{\sqrt{{\displaystyle 1-\frac{1}{\mathrm{x}}}}+1} \\ =2-\frac{1}{2} \\ =\frac{3}{2} \end{gathered}$$