1. **Problem Statement:** We want to find the limit $$\lim_{x \to 4} \frac{x^3 - 2x^2 - 9x + 4}{3x^2 - 4x}$$ 2. **Formula and Important Rules:** When direct substitution in a limit results in an indeterminate form like $\frac{0}{0}$, we try to simplify the expression by factoring or algebraic manipulation. 3. **Step-by-step Solution:** - Substitute $x=4$ directly: $$\frac{4^3 - 2 \cdot 4^2 - 9 \cdot 4 + 4}{3 \cdot 4^2 - 4 \cdot 4} = \frac{64 - 32 - 36 + 4}{48 - 16} = \frac{0}{0}$$ - Since this is indeterminate, factor numerator and denominator: Numerator: $x^3 - 2x^2 - 9x + 4$ Try to factor by grouping or synthetic division. Since $x=4$ is a root, $(x-4)$ is a factor. Divide numerator by $(x-4)$: $$\frac{x^3 - 2x^2 - 9x + 4}{x - 4} = x^2 + 2x - 1$$ Denominator: $3x^2 - 4x = x(3x - 4)$ Rewrite original limit: $$\lim_{x \to 4} \frac{(x - 4)(x^2 + 2x - 1)}{x(3x - 4)}$$ Cancel $(x - 4)$: $$\lim_{x \to 4} \frac{x^2 + 2x - 1}{3x - 4}$$ - Substitute $x=4$ now: $$\frac{4^2 + 2 \cdot 4 - 1}{3 \cdot 4 - 4} = \frac{16 + 8 - 1}{12 - 4} = \frac{23}{8}$$ 4. **Final Answer:** $$\boxed{\frac{23}{8}}$$ This means the limit exists and equals $\frac{23}{8}$. If you want, I can explain any step in more detail!