Ответ:
0
Пошаговое объяснение:
Сделаем замену: 
=> 
\pi} \frac{1+\cos x}{x-\pi} = \lim_{y->0} \frac{1+\cos(y+\pi)}{y} = \lim_{y->0} \frac{1-\cos y}{y} = \lim_{y->0} \frac{1-(1-2\sin^2 \frac{y}{2})}{y} = \lim_{y->0} \frac{\sin^2 \frac{y}{2}}{y/2} = \lim_{y->0} (\frac{\sin \frac{y}{2}}{y/2})^2 \cdot \frac{y}{2} = \lim_{y->0} (\frac{\sin \frac{y}{2}}{y/2})^2 \cdot \lim_{y->0} \frac{y}{2} = 1^2 \cdot 0 = 0." alt="\lim_{x->\pi} \frac{1+\cos x}{x-\pi} = \lim_{y->0} \frac{1+\cos(y+\pi)}{y} = \lim_{y->0} \frac{1-\cos y}{y} = \lim_{y->0} \frac{1-(1-2\sin^2 \frac{y}{2})}{y} = \lim_{y->0} \frac{\sin^2 \frac{y}{2}}{y/2} = \lim_{y->0} (\frac{\sin \frac{y}{2}}{y/2})^2 \cdot \frac{y}{2} = \lim_{y->0} (\frac{\sin \frac{y}{2}}{y/2})^2 \cdot \lim_{y->0} \frac{y}{2} = 1^2 \cdot 0 = 0." align="absmiddle" class="latex-formula">