1. Discuss continuity of $f(x) = \begin{cases} \frac{x^2 - 9}{x - 3}, & x \neq 3 \\ 0, & x = 3 \end{cases}$ at $x=3$. Simplify the function for $x \neq 3$: $$\frac{x^2 - 9}{x - 3} = \frac{(x - 3)(x + 3)}{x - 3} = x + 3, \quad x \neq 3.$$ So, the limit as $x \to 3$ is: $$\lim_{x \to 3} f(x) = \lim_{x \to 3} (x + 3) = 6.$$ The function value at $x=3$ is $f(3) = 0$. Since $$\lim_{x \to 3} f(x) \neq f(3),$$ $f$ is not continuous at $x=3$. 2. Discuss continuity of $f(x) = \begin{cases} x - 4, & -1 < x \leq 2 \\ x^2 - 6, & 2 < x < 5 \end{cases}$ at $x=2$. Calculate left-hand limit (LHL): $$\lim_{x \to 2^-} f(x) = 2 - 4 = -2.$$ Calculate right-hand limit (RHL): $$\lim_{x \to 2^+} f(x) = 2^2 - 6 = 4 - 6 = -2.$$ Function value at $x=2$ is from left piece since $2 \leq 2$: $$f(2) = 2 - 4 = -2.$$ Since LHL = RHL = $f(2)$, $f$ is continuous at $x=2$. 3. Discuss continuity of $f(x) = \begin{cases} x \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x = 0 \end{cases}$ at $x=0$. Calculate the limit as $x \to 0$: Since $|\sin(1/x)| \leq 1$, $$|x \sin(1/x)| \leq |x|,$$ and $$\lim_{x \to 0} |x| = 0,$$ thus by the squeeze theorem, $$\lim_{x \to 0} x \sin\left(\frac{1}{x}\right) = 0.$$ The function value at 0 is $f(0) = 0$, so $f$ is continuous at $x=0$. 4. Find $c$ such that $f(x) = \begin{cases} x, & x \text{ is irrational} \\ 1 - x, & x \text{ is rational} \end{cases}$ is continuous on $[0,1]$, in particular at $x=\frac{1}{2}$. For continuity at $x=\frac{1}{2}$, the limit from rationals and irrationals must be equal: $$\lim_{x \to 1/2 \text{ rational}} f(x) = 1 - \frac{1}{2} = \frac{1}{2},$$ $$\lim_{x \to 1/2 \text{ irrational}} f(x) = \frac{1}{2}.$$ Both limits are $\frac{1}{2}$, so to make $f$ continuous at $x=\frac{1}{2}$, define: $$f\left(\frac{1}{2}\right) = c = \frac{1}{2}.$$ 5. Discuss continuity of $f(x) = \begin{cases} x - 4, & x \geq 0, x \neq 4 \end{cases}$ at $x=4$. Here, $f$ is not defined at $x=4$, so to discuss continuity, we must define $f(4)$. Assuming $f(4) = c$, continuity requires: $$\lim_{x \to 4} f(x) = f(4) = c.$$ Compute limit: Since $f(x) = x - 4$ near $4$, $$\lim_{x \to 4} f(x) = 4 - 4 = 0,$$ so $$c = 0$$ for continuity at $x=4$. Final answers: - $f$ not continuous at $x=3$. - $f$ continuous at $x=2$. - $f$ continuous at $x=0$. - $c = \frac{1}{2}$ for continuity at $x=\frac{1}{2}$. - $c = 0$ for continuity at $x=4$.