ข้อสอบ A-Level คณิตประยุกต์ 1 ปี 2559 (วิชาสามัญเก่า)

$$\begin{gathered} \mathrm{จากโจทย์}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \underset{\mathrm{x}\to 2}{\lim }\left(\frac{2}{\mathrm{x}-2}+\frac{1}{\mathrm{x}+2}-\frac{8}{{\mathrm{x}}^2-4}\right) \\ \mathrm{จะได้} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\underset{\mathrm{x}\to 2}{\lim }\left(\frac{2}{\mathrm{x}-2}+\frac{1}{\mathrm{x}+2}-\frac{8}{\left(\mathrm{x}-2\right)\left(\mathrm{x}+2\right)}\right) \\ = \underset{\mathrm{x}\to 2}{\lim }\left(\frac{2\left(\mathrm{x}+2\right)+1\left(\mathrm{x}-2\right)-8}{\left(\mathrm{x}-2\right)\left(\mathrm{x}+2\right)}\right) \\ =\underset{\mathrm{x}\to 2}{\lim }\left(\frac{2\left(\mathrm{x}+2\right)+1\left(\mathrm{x}-2\right)-8}{\left(\mathrm{x}-2\right)\left(\mathrm{x}+2\right)}\right) \end{gathered}$$

$$\begin{gathered} =\underset{\mathrm{x}\to 2}{\lim }\frac{(3\mathrm{x}-6)}{(\mathrm{x}-2)(\mathrm{x}+2)} \\ =\underset{\mathrm{x}\to 2}{\lim }\frac{3\cancel{(\mathrm{x}-2)}}{\left(\cancel{\mathrm{x}-2}\right)(\mathrm{x}+2)} \\ =\underset{\mathrm{x}\to 2}{\lim }\left(\frac{3}{\mathrm{x}+2}\right)\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}; \mathrm{แทน} \mathrm{x}=2 \\ =\frac{3}{2+2} \\ =\frac{3}{4} \end{gathered}$$