1. **Problem statement:** Find the limit $$\lim_{x \to 2} \frac{x^4 - 8x}{\frac{1}{4}x^3 + 5x - 6x^2}$$ using L'Hopital's rule. 2. **Check the form:** Substitute $x=2$: $$\text{Numerator} = 2^4 - 8 \times 2 = 16 - 16 = 0$$ $$\text{Denominator} = \frac{1}{4} \times 2^3 + 5 \times 2 - 6 \times 2^2 = \frac{1}{4} \times 8 + 10 - 24 = 2 + 10 - 24 = -12$$ Since denominator is $-12 \neq 0$, the limit is of the form $\frac{0}{-12} = 0$. 3. **Conclusion:** The limit is simply 0 because the numerator approaches 0 and denominator approaches $-12$ (non-zero). **Final answer:** $$\boxed{0}$$