1. **State the problem:** We have the piecewise function $$f(x)=\begin{cases}\sqrt{-x} & \text{if } x<0\\3 - x & \text{if } 0 \leq x < 3\\(x - 3)^2 & \text{if } x \geq 3\end{cases}$$ We want to: (i) Find $\lim_{x\to 0} f(x)$ if it exists. (ii) Show that $f(x)$ is continuous at $x=3$. (iii) Sketch the graph of $f(x)$. 2. **Find $\lim_{x\to 0} f(x)$:** - Left-hand limit as $x\to 0^-$: Since $x<0$, use $f(x) = \sqrt{-x}$. $$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \sqrt{-x} = \sqrt{0} = 0.$$ - Right-hand limit as $x\to 0^+$: Since $0 \leq x < 3$, use $f(x) = 3-x$. $$\lim_{x \to 0^+} f(x) = 3 - 0 = 3.$$ - Since $\lim_{x \to 0^-} f(x) = 0 \neq 3 = \lim_{x \to 0^+} f(x),$ the limit $\lim_{x \to 0} f(x)$ does not exist. 3. **Show continuity at $x=3$:** - Compute $f(3)$ using the third case $f(x) = (x-3)^2$ for $x \geq 3$: $$f(3) = (3 - 3)^2 = 0.$$ - Left-hand limit as $x \to 3^-$, use $f(x) = 3 - x$: $$\lim_{x \to 3^-} f(x) = 3 - 3 = 0.$$ - Right-hand limit as $x \to 3^+$, use $f(x) = (x - 3)^2$: $$\lim_{x \to 3^+} f(x) = (3 - 3)^2 = 0.$$ - Both limits and the function value at 3 are equal: $$\lim_{x \to 3} f(x) = f(3) = 0,$$ so, $f$ is continuous at $x=3$. 4. **Sketch the graph:** - For $x<0$, graph $y = \sqrt{-x}$, which is the square root reflected over the y-axis. - For $0 \leq x < 3$, graph $y = 3 - x$, a line decreasing from 3 at $x=0$ to 0 at $x=3$. - For $x \geq 3$, graph $y = (x - 3)^2$, a parabola starting at 0 at $x=3$ and opening upwards. **Final answers:** (i) $\lim_{x \to 0} f(x)$ does not exist. (ii) $f$ is continuous at $x=3$. (iii) The graph is as described above.