Subtitusi langsung $x=\pi$ menghasilkan bentuk tak tentu $\frac{0}{0}$.

Ingat bahwa

$\cos2x=\cos^2x-\sin^2x$

Dengan demikian, diperoleh

$\lim\limits_{x\rightarrow\pi}\ \frac{\cos2x}{\sin x-\cos x}=\lim\limits_{x\rightarrow\pi}\ \left(\frac{\cos2x}{\sin x-\cos x}\cdot\frac{\sin x+\cos x}{\sin x+\cos x}\ \right)$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\lim\limits_{x\rightarrow\pi}\ \frac{\cos2x\left(\sin x+\cos x\right)}{\sin^2x-\cos^2x}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\lim\limits_{x\rightarrow\pi}\ \frac{\cos2x\left(\sin x+\cos x\right)}{-\cos2x}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\lim\limits_{x\rightarrow\pi}\ -\left(\sin x+\cos x\right)$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ -\left(\sin\pi+\cos\pi\right)$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ -\left(0+\left(-1\right)\right)$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ 1$

Jadi, nilai $\lim\limits_{x\rightarrow\pi}\ \frac{\cos2x}{\sin x-\cos x}=1$