\begin{aligned} \lim_{x \to 0} \ \frac{x \tan 5x}{\cos 2x - \cos 7x} &= \lim_{x \to 0} \ \frac{x \tan 5x}{-2 \sin \left( \frac{2x+7x}{2} \right) \sin \left( \frac{2x-7x}{2} \right)} \\[8pt] &= \lim_{x \to 0} \ \frac{x \tan 5x}{-2 \sin \left( \frac{9}{2}x \right) \sin \left( -\frac{5}{2} x \right)} \\[8pt] &= \lim_{x \to 0} \ \frac{x \tan 5x}{-2 \sin \left( \frac{9}{2}x \right) \left( -\sin (\frac{5}{2} x) \right)} \\[8pt] &= \lim_{x \to 0} \ \frac{x \tan 5x}{2 \sin \left( \frac{9}{2}x \right) \left( \sin (\frac{5}{2} x) \right)} \\[8pt] &= \frac{1}{2} \cdot \lim_{x \to 0} \ \frac{x}{\sin \left( \frac{9}{2}x \right)} \cdot \lim_{x \to 0} \ \frac{\tan 5x}{\sin (\frac{5}{2} x)} \\[8pt] &= \frac{1}{2} \cdot \frac{1}{\frac{9}{2}} \cdot \frac{5}{\frac{5}{2}} = \frac{1}{2} \cdot \frac{2}{9} \cdot 2 \\[8pt] &= \frac{2}{9} \end{aligned}