1. The problem asks to sketch a graph of a function $f(x)$ illustrating the given limits and function values for problem 9. 2. Given: - $\lim_{x \to 1} f(x)$ does not exist (DNE). - $\lim_{x \to 0} f(x) = 3$. - $f(-1) = -3$. - $f(x) = 3$ if $x < 1$. - Domain: all real numbers except 0. 3. Important rules: - The limit $\lim_{x \to c} f(x)$ exists only if the left-hand and right-hand limits at $c$ are equal. - If $f(c)$ is defined and different from the limit, the graph has a hole at $(c, \lim_{x \to c} f(x))$ and a filled circle at $(c, f(c))$. - If the limit does not exist at $c$, the left and right limits differ or do not exist. 4. For $x < 1$, $f(x) = 3$ is constant, so the graph is a horizontal line at $y=3$ for $x<1$. 5. Since $\lim_{x \to 1} f(x)$ does not exist, the left and right limits at $x=1$ differ or one side is undefined. 6. At $x=0$, $\lim_{x \to 0} f(x) = 3$, but $0$ is not in the domain, so there is a vertical asymptote or hole at $x=0$. 7. At $x=-1$, $f(-1) = -3$, so the graph passes through $(-1, -3)$. 8. Sketch summary: - Horizontal line $y=3$ for $x<1$. - At $x=1$, the limit does not exist, so the graph jumps or breaks. - At $x=0$, no point, but limit 3 from both sides. - Point at $(-1, -3)$. Final answer: The graph is a horizontal line at $y=3$ for $x<1$ with a jump or break at $x=1$ (limit DNE), a hole or asymptote at $x=0$ with limit 3, and a point at $(-1, -3)$.