1. **Problem (I):** Calculate $$\lim_{x \to 0} \frac{27 - (3 + x)^3}{x}$$ 2. **Step 1:** Expand the cube in the numerator: $$ (3 + x)^3 = 3^3 + 3 \cdot 3^2 \cdot x + 3 \cdot 3 \cdot x^2 + x^3 = 27 + 27x + 9x^2 + x^3 $$ 3. **Step 2:** Substitute back: $$ \frac{27 - (27 + 27x + 9x^2 + x^3)}{x} = \frac{27 - 27 - 27x - 9x^2 - x^3}{x} = \frac{-27x - 9x^2 - x^3}{x} $$ 4. **Step 3:** Simplify by dividing numerator and denominator by $$x$$: $$ \frac{-27x - 9x^2 - x^3}{x} = -27 - 9x - x^2 $$ 5. **Step 4:** Take the limit as $$x \to 0$$: $$ \lim_{x \to 0} (-27 - 9x - x^2) = -27 $$ --- 6. **Problem (II):** Calculate $$\lim_{x \to 4} \frac{3x^2 - 11x - 7}{5x^2 - 19x - 4}$$ 7. **Step 1:** Factor numerator and denominator if possible. - Numerator: $$3x^2 - 11x - 7$$ Try to factor: Find two numbers that multiply to $$3 \times (-7) = -21$$ and add to $$-11$$. These are $$-7$$ and $$3$$. Rewrite: $$3x^2 - 7x + 3x - 7 = (3x^2 - 7x) + (3x - 7) = x(3x - 7) + 1(3x - 7) = (x + 1)(3x - 7)$$ - Denominator: $$5x^2 - 19x - 4$$ Find two numbers that multiply to $$5 \times (-4) = -20$$ and add to $$-19$$. These are $$-20$$ and $$1$$. Rewrite: $$5x^2 - 20x + x - 4 = (5x^2 - 20x) + (x - 4) = 5x(x - 4) + 1(x - 4) = (5x + 1)(x - 4)$$ 8. **Step 2:** Substitute factored forms: $$ \frac{(x + 1)(3x - 7)}{(5x + 1)(x - 4)} $$ 9. **Step 3:** Evaluate the limit by direct substitution $$x = 4$$: Numerator: $$ (4 + 1)(3 \cdot 4 - 7) = 5 \times (12 - 7) = 5 \times 5 = 25 $$ Denominator: $$ (5 \cdot 4 + 1)(4 - 4) = (20 + 1) \times 0 = 21 \times 0 = 0 $$ Since denominator approaches zero and numerator approaches 25, the limit does not exist as a finite number. 10. **Step 4:** Check the behavior near $$x = 4$$: - For $$x \to 4^-$$ (values slightly less than 4), $$x - 4 < 0$$, denominator is negative small number times positive number, so denominator is negative small. - For $$x \to 4^+$$ (values slightly greater than 4), $$x - 4 > 0$$, denominator is positive small. Numerator near 4 is positive (25). Therefore: - $$\lim_{x \to 4^-} \frac{3x^2 - 11x - 7}{5x^2 - 19x - 4} = -\infty$$ - $$\lim_{x \to 4^+} \frac{3x^2 - 11x - 7}{5x^2 - 19x - 4} = +\infty$$ 11. **Final conclusion:** The limit does not exist because the left and right limits are infinite and not equal. --- **Final answers:** (I) $$-27$$ (II) Limit does not exist (infinite discontinuity at $$x=4$$).