Why can't I?

Let $f(x) = x\cos x$ , then $\displaystyle \lim_{x \to \pi} \dfrac{x\cos x - (-\pi)}{x-\pi} = \lim_{x \to \pi} \dfrac{f(x) - f(\pi)}{x-\pi}= f'(\pi) = \cos \pi - \pi \sin \pi = \boxed{-1}$

$\large \lim_{x \to \pi} \frac {\pi + x\cos(x)}{ x - \pi} = lim_{x \to \pi} \frac {\pi - x}{x - \pi} = \boxed{-1}$

Using Taylor serie for $\cos x$ around $x = \pi$

$\lim_{x \to \pi}\frac{\pi+x\cos(x)}{x-\pi}×\frac{\pi-x\cos(x)}{\pi-x\cos(x)}$ $=>$ $\lim_{x\to \pi}\frac{\pi^{2}-x^{2}\cos^{2}(x)}{(x-\pi)(\pi-x\cos(x))}$ $=> \lim_{x \to \pi} \frac{\pi^{2}-x^{2}+x^{2}-x^{2}\cos^{2}(x)}{(x-\pi)(\pi-x\cos(x))}$ $=\lim_{x \to \pi}\frac{\pi^{2}-x^{2}}{(x-\pi)(\pi-x\cos(x))}+\lim_{x \to \pi}\frac{x^{2}(1-\cos^{2}(x)}{(x-\pi)(\pi-x\cos(x)}$ $=\lim_{x \to \pi} \frac{-\boxed{x-\pi}(x+\pi)}{\boxed{x-\pi}(\pi-x\cos(x))}+\lim_{x \to \pi}\frac{x^{2}\sin(x)\sin(x)}{(x-\pi)(\pi-x\cos(x))}$ $=\frac{-2\pi}{2\pi}+\lim_{x \to \pi}\frac{\sin(x)}{x- \pi}×\lim_{x \to \pi} \frac{x^{2}\sin(x)}{\pi-x\cos(x)}$ $=-1+\lim_{x \to \pi} \frac{\sin(\pi-x)}{-(\pi-x)}×(0)$ $=-1+(-1×0)$ $=\boxed{-1}$