1. **State the problem:** Evaluate the limit $$\lim_{(x,y)\to(2,-4)} \frac{x+4}{x^2y - xy + 4^2 - 4x}$$ by factorization. 2. **Substitute the point** $(2,-4)$ directly to check for indeterminate form: $$\frac{2+4}{2^2 (-4) - 2(-4) + 16 - 4\cdot2} = \frac{6}{4(-4) + 8 + 16 - 8} = \frac{6}{-16 + 8 + 16 - 8} = \frac{6}{0}$$ The denominator is zero, so direct substitution is undefined. 3. **Rewrite and factorize the denominator:** Denominator: $$x^2 y - x y + 4^2 - 4x = yx^2 - yx + 16 - 4x$$ Group terms: $$y x^2 - y x - 4 x + 16$$ Factor $x$ from the first three terms: $$x^2 y - x y - 4 x + 16 = x(y x - y - 4) + 16$$ 4. **Check if further factorization is possible:** Rewrite $y x - y - 4$ as: $$y(x-1) - 4$$ So denominator: $$x [y(x-1) -4] + 16$$ 5. **Evaluate the denominator near (2,-4):** Substitute $x=2$ and $y=-4$: $$2[-4(2-1) -4] + 16 = 2[-4(1) -4] + 16 = 2[-4 -4] +16 = 2[-8] +16 = -16 +16 = 0$$ Still zero. 6. **Try to simplify numerator and denominator expressions:** Numerator at $(2,-4)$ is $6$, denominator is $0$. To check if the limit exists, analyze limit from different paths. 7. **Use path approach:** Let $y = mx + b$ approach $(2,-4)$; check limit along paths to see if limit depends on path: Try $y = -4$ (constant): Denominator: $x^2(-4) - x(-4) +16 -4x = -4 x^2 + 4 x +16 - 4 x = -4 x^2 +16$ At $x \to 2$: denomin= $-4(4)+16 = -16 +16 =0$ Numerator $x+4 = 6$ So limit $ o \frac{6}{0}$ (which diverges). Try $y = \frac{4(x-1)}{x}$: This is derived from $y(x-1)-4=0$ to make denominator zero at the algebraic expression inside; this path makes denominator simplifying to $16$? Verify. Since denominator tends to zero and numerator tends to a nonzero number, the limit does not exist or is infinite. **Final conclusion:** The limit $$\lim_{(x,y)\to(2,-4)} \frac{x+4}{x^2 y - x y + 16 - 4x}$$ does not exist because the denominator approaches zero while numerator approaches 6, leading to an unbounded expression. **Answer:** The limit does not exist.