ข้อสอบ PAT 1 - มีนาคม 2555

$$\begin{gathered} \mathrm{จากโจทย์} \underset{\mathrm{x}\to \frac{\mathrm{\pi }}{4}}{\lim }\frac{({\cot }^3\mathrm{x}-1){\mathrm{cosec}}^2\mathrm{x}}{1+\cos 2\mathrm{x}-2{\sin }^2\mathrm{x}} \\ =\underset{\mathrm{x}\to \frac{\mathrm{\pi }}{4}}{\lim }\frac{({\cot }^3\mathrm{x}-1^3){\mathrm{cosec}}^2\mathrm{x}}{1+\cos 2\mathrm{x}-2{\sin }^2\mathrm{x}} \\ =\underset{\mathrm{x}\to \frac{\mathrm{\pi }}{4}}{\lim }\frac{(\mathrm{cotx}-1)\left({\cot }^2\mathrm{x}+\mathrm{cotx}+1\right){\mathrm{cosec}}^2\mathrm{x}}{1+\left(2{\cos }^2\mathrm{x}-1\right)-2{\sin }^2\mathrm{x}} \\ \mathrm{คูณ} \frac{\mathrm{sinx}}{\mathrm{sinx}} \mathrm{ทั้งสมการ} \mathrm{จะได้} \end{gathered}$$
$$\begin{gathered} =\underset{\mathrm{x}\to \frac{\mathrm{\pi }}{4}}{\lim }\left(\frac{\mathrm{sinx}}{\mathrm{sinx}}\right)·\frac{({\displaystyle \frac{\mathrm{cosx}}{\mathrm{sinx}}}-1)\left({\cot }^2\mathrm{x}+\mathrm{cotx}+1\right){\mathrm{cosec}}^2\mathrm{x}}{\left(2{\cos }^2\mathrm{x}-2{\sin }^2\mathrm{x}\right)} \\ =\underset{\mathrm{x}\to \frac{\mathrm{\pi }}{4}}{\lim }\frac{\cancel{(\cos x-\sin x)}\left({\cot }^2\mathrm{x}+\mathrm{cotx}+4\right){\mathrm{cosec}}^2\mathrm{x}}{2\sin x\cancel{(\cos x-\sin x)}\left(\mathrm{cosx}+\mathrm{sinx}\right)} \\ =\underset{\mathrm{x}\to \frac{\mathrm{\pi }}{4}}{\lim }\frac{\left({\cot }^2\mathrm{x}+\mathrm{cotx}+4\right){\mathrm{cosec}}^2\mathrm{x}}{2\sin x\left(\mathrm{cosx}+\mathrm{sinx}\right)} \end{gathered}$$

$$\begin{gathered} =\frac{\left({\cot }^2{\displaystyle \frac{\mathrm{\pi }}{4}}+\cot {\displaystyle \frac{\mathrm{\pi }}{4}}+4\right){\mathrm{cosec}}^2{\displaystyle \frac{\mathrm{\pi }}{4}}}{\left(2\sin \frac{\mathrm{\pi }}{4}\right)\left(\cos \frac{\mathrm{\pi }}{4}+\sin \frac{\mathrm{\pi }}{4}\right)} \\ =\frac{\left(1+1+1\right)\left(2\right)}{2\left({\displaystyle \frac{\sqrt{2}}{2}}\right)\left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\right)} \\ =3 \end{gathered}$$