1. **Evaluate the limit** $$\lim_{x \to 7} \frac{7x^{2} - 40x - 63}{x - 7}$$. Step 1: Factor the numerator if possible. Consider the quadratic expression in the numerator: $$7x^{2} - 40x - 63$$. Step 2: Factor out 7: $$7x^{2} - 40x - 63 = 7x^{2} - 40x - 63$$ (40 is not divisible by 7, so factorization is not straightforward). Instead, use polynomial division or evaluate the limit by simplifying the expression. Step 3: Since direct substitution results in division by zero (denominator $$x-7=0$$), try to factor numerator to cancel $$x-7$$. Evaluate numerator at $$x=7$$: $$7(7)^{2} - 40(7) - 63 = 7(49) - 280 - 63 = 343 - 280 - 63 = 0$$ So, $$x=7$$ is root of numerator, hence $$x-7$$ is a factor. Step 4: Perform polynomial division of numerator by $$x - 7$$: Divide $$7x^{2} - 40x - 63$$ by $$x - 7$$: - First term: $$7x^{2} \div x = 7x$$. - Multiply: $$7x(x-7) = 7x^{2} - 49x$$. - Subtract: $$(7x^{2} - 40x - 63) - (7x^{2} - 49x) = 9x - 63$$. - Next term: $$9x \div x = 9$$. - Multiply: $$9(x -7) = 9x - 63$$. - Subtract: $$(9x - 63) - (9x - 63) = 0$$. So quotient is $$7x + 9$$. Step 5: Thus, $$\frac{7x^{2} - 40x - 63}{x - 7} = 7x + 9$$ for $$x \neq 7$$. Step 6: Take the limit as $$x \to 7$$: $$\lim_{x \to 7} (7x + 9) = 7(7) + 9 = 49 + 9 = 58$$. 2. **Test the continuity of the function** $$f(x) = \begin{cases} \frac{8}{\sqrt{4}} x - 1, & x < 1 \\ \frac{7}{\sqrt[3]{x}} \cdot \sqrt{x} - 4, & 1 \le x \le 2 \\ -5x - \frac{4}{3}, & x > 2 \end{cases}$$ **at** $$x=1$$ and $$x=2$$. --- At $$x=1$$: Step 1: Compute $$f(1)$$ from the middle piece since $$1 \le x \le 2$$: Evaluate: $$f(1) = \frac{7}{\sqrt[3]{1}} \cdot \sqrt{1} - 4 = 7 \cdot 1 - 4 = 3$$. Step 2: Find left-hand limit as $$x \to 1^-$$: $$f(x) = \frac{8}{\sqrt{4}} x - 1 = \frac{8}{2}x -1 = 4x -1$$. So, $$\lim_{x \to 1^-} f(x) = 4(1) -1 = 3$$. Step 3: Find right-hand limit as $$x \to 1^+$$: The function expression for $$1 \le x \le 2$$ is: $$f(x) = \frac{7}{\sqrt[3]{x}} \cdot \sqrt{x} - 4$$. Substitute $$x=1$$: $$\lim_{x \to 1^+} f(x) = 7 \cdot 1 - 4 = 3$$. Step 4: Since $$f(1) = 3 = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x)$$, function is continuous at $$x=1$$. --- At $$x=2$$: Step 1: Compute $$f(2)$$ from middle piece: Evaluate: $$f(2) = \frac{7}{\sqrt[3]{2}} \cdot \sqrt{2} -4$$. Since $$\sqrt[3]{2} = 2^{1/3}$$ and $$\sqrt{2} = 2^{1/2}$$, $$f(2) = 7 \cdot \frac{2^{1/2}}{2^{1/3}} - 4 = 7 \cdot 2^{(1/2 - 1/3)} - 4 = 7 \cdot 2^{(3/6 - 2/6)} -4 = 7 \cdot 2^{1/6} -4$$. Step 2: Find left-hand limit as $$x \to 2^-$$: Same as $$f(2)$$ above, since left side of 2 is within $$1 \le x \le 2$$ piece, so $$\lim_{x \to 2^-} f(x) = f(2) = 7 \cdot 2^{1/6} -4$$. Step 3: Find right-hand limit as $$x \to 2^+$$: For $$x > 2$$, $$f(x) = -5x - \frac{4}{3}$$. Evaluate limit: $$\lim_{x \to 2^+} f(x) = -5(2) - \frac{4}{3} = -10 - \frac{4}{3} = -\frac{30}{3} - \frac{4}{3} = -\frac{34}{3} \approx -11.3333$$. Step 4: Compare left and right limits: $$\lim_{x \to 2^-} f(x) = 7 \cdot 2^{1/6} -4 \approx 7 \cdot 1.122 -4 = 7.854 -4 = 3.854$$ and $$\lim_{x \to 2^+} f(x) = -11.3333$$ These are not equal, so the limit does not exist at 2. Step 5: Hence, function is **not continuous** at $$x=2$$ because right and left limits are not equal. --- **Final answers:** 1. $$\lim_{x \to 7} \frac{7x^{2} - 40x - 63}{x - 7} = 58$$. 2. Function $$f(x)$$ is continuous at $$x=1$$ and discontinuous at $$x=2$$.