1. The problem asks to complete the table by finding either the limit $\lim_{x \to c} f(x)$ or the function value $f(c)$ for each given function and point $c$. 2. Recall that if $f$ is continuous at $x=c$, then $\lim_{x \to c} f(x) = f(c)$. For rational and polynomial functions, limits can often be found by direct substitution. 3. For the first row: $f(x) = x^2 - 2$, $c=0$, and $f(c) = f(0) = -2$ is given. Find $\lim_{x \to 0} f(x)$ by substitution: $$\lim_{x \to 0} (x^2 - 2) = 0^2 - 2 = -2$$ 4. For the second row: $f(x) = (3x - 4)^2$, $c = -1$, and $f(c) = 49$ is given. Find $\lim_{x \to -1} f(x)$ by substitution: $$\lim_{x \to -1} (3x - 4)^2 = (3(-1) - 4)^2 = (-3 - 4)^2 = (-7)^2 = 49$$ 5. For the third row: $f(x) = \frac{x + 1}{x}$, $c=1$, and $\lim_{x \to 1} f(x) = 2$ is given. Find $f(1)$: $$f(1) = \frac{1 + 1}{1} = \frac{2}{1} = 2$$ 6. For the fourth row: $f(x) = \sqrt{3x - 5}$, $c=2$, and $\lim_{x \to 2} f(x) = \pm 1$ is given. Find $f(2)$: $$f(2) = \sqrt{3(2) - 5} = \sqrt{6 - 5} = \sqrt{1} = 1$$ 7. Note that the limit $\pm 1$ suggests the limit might be $1$ since the square root function outputs the principal (non-negative) root. Final completed table values: - Row 1 limit: $-2$ - Row 2 limit: $49$ - Row 3 function value: $2$ - Row 4 function value: $1$