จงหาค่าของlimx→−π4(sec2x)(1+cot3x)2sin2x−cos2x−1
,
ทดลองแทนค่า $x=-\frac{\pi}{4}$ พบว่า อยู่ในรูป $\frac{0}{0}$
แปลงให้อยู่ในรูป $\sin$ กับ $\cos$ และใช้สูตรมุมสองเท่า $\cos{2x}=2\cos^2{x}-1$
$$\lim_{x\rightarrow-\frac{\pi}{4}}\frac{\left(\frac{1}{\cos^2{x}}\right)\left(\frac{\sin^3{x}}{\sin^3{x}}+\frac{\cos^3{x}}{\sin^3{x}}\right)}{2\sin^2{x}-(2\cos^2{x}-1)-1}=\lim_{x\rightarrow-\frac{\pi}{4}}\frac{\sin^3{x}+\cos^3{x}}{2(\cos^2{x})(\sin^3{x})(\sin^2{x}-\cos^2{x})}$$
,
\begin{eqnarray*}
\sin^3{x}+\cos^3{x}&=&(\sin{x}+\cos{x})(\sin^2{x}-(\sin{x})(\cos{x})+\cos^2{x})\\
\sin^2{x}-\cos^2{x}&=&(\sin{x}-\cos{x})(\sin{x}+\cos{x})
\end{eqnarray*}
ตัดตัวประกอบร่วมกันออกไป
$$\lim_{x\rightarrow-\frac{\pi}{4}}\frac{(\sin{x}+\cos{x})(\sin^2{x}-(\sin{x})(\cos{x})+\cos^2{x})}{2(\cos^2{x})(\sin^3{x})(\sin{x}-\cos{x})(\sin{x}+\cos{x})}=\lim_{x\rightarrow-\frac{\pi}{4}}\frac{\sin^2{x}-(\sin{x})(\cos{x})+\cos^2{x}}{2(\cos^2{x})(\sin^3{x})(\sin{x}-\cos{x}} $$
,
${\displaystyle\lim_{x\rightarrow-\frac{\pi}{4}}\frac{\left(\sec^{2}x\right)\left(1+\cot^{3}x\right)}{2\sin^{2}x-\cos2x-1}}$
\begin{eqnarray*}
& = & \lim_{x\rightarrow-\frac{\pi}{4}}\frac{\sin^{2}x-\left(\sin x\right)\left(\cos x\right)+\cos^{2}x}{2\left(\cos^{2}x\right)\left(\sin^{3}x\right)\left(\sin x-\cos x\right)}\\
& = & \lim_{x\rightarrow-\frac{\pi}{4}}\frac{1-\sin x\cos x}{2\cos^{2}x\sin^{3}x\left(\sin x-\cos x\right)}\\
& = & \frac{1-\left(-\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)}{2\left(\frac{1}{\sqrt{2}}\right)^{2}\left(-\frac{1}{\sqrt{2}}\right)^3\left(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right)}\\
& = & 3
\end{eqnarray*}