1. We are asked to find the limit: $$\lim_{x \to \frac{\pi}{8}} \frac{\sen \left( \frac{5\pi}{4} - 2x \right)}{2x - \frac{\pi}{4}}$$ 2. Recall that $\sen$ is the sine function. The limit involves a trigonometric expression in the numerator and a linear expression in the denominator. 3. First, evaluate the denominator at $x = \frac{\pi}{8}$: $$2 \cdot \frac{\pi}{8} - \frac{\pi}{4} = \frac{\pi}{4} - \frac{\pi}{4} = 0$$ 4. Evaluate the numerator at $x = \frac{\pi}{8}$: $$\sen \left( \frac{5\pi}{4} - 2 \cdot \frac{\pi}{8} \right) = \sen \left( \frac{5\pi}{4} - \frac{\pi}{4} \right) = \sen \left( \pi \right) = 0$$ 5. Since both numerator and denominator approach 0, we have an indeterminate form $\frac{0}{0}$, so we can apply L'Hôpital's Rule. 6. Differentiate numerator and denominator with respect to $x$: - Numerator derivative: $$\frac{d}{dx} \sen \left( \frac{5\pi}{4} - 2x \right) = \cos \left( \frac{5\pi}{4} - 2x \right) \cdot (-2) = -2 \cos \left( \frac{5\pi}{4} - 2x \right)$$ - Denominator derivative: $$\frac{d}{dx} \left( 2x - \frac{\pi}{4} \right) = 2$$ 7. Applying L'Hôpital's Rule, the limit becomes: $$\lim_{x \to \frac{\pi}{8}} \frac{-2 \cos \left( \frac{5\pi}{4} - 2x \right)}{2}$$ 8. Simplify the fraction by canceling 2: $$\frac{\cancel{-2} \cos \left( \frac{5\pi}{4} - 2x \right)}{\cancel{2}} = - \cos \left( \frac{5\pi}{4} - 2x \right)$$ 9. Evaluate the limit by substituting $x = \frac{\pi}{8}$: $$- \cos \left( \frac{5\pi}{4} - 2 \cdot \frac{\pi}{8} \right) = - \cos \left( \frac{5\pi}{4} - \frac{\pi}{4} \right) = - \cos (\pi) = - (-1) = 1$$ 10. Therefore, the limit is: $$\boxed{1}$$