Continuity

Root in a denominator: $e^{2x} - 4 > 0$. Log: $6 - x > 0$.

$e^{2x} > 4 \Rightarrow 2x > \log 4 \Rightarrow x > \tfrac{1}{2}\log 4 = \log 2$. And $6 - x > 0 \Rightarrow x < 6$.

Both at once: $\log 2 < x < 6$.

$(\log 2,\ 6)$.

Even root on top: $x + 4 \geq 0$. Denominator: $x^2 - x - 6 \neq 0$.

$x \geq -4$. And $x^2 - x - 6 = (x - 3)(x + 2) \neq 0 \Rightarrow x \neq 3,\ x \neq -2$.

$x \geq -4$ with the two points $-2$ and $3$ removed.

$[-4, -2) \cup (-2, 3) \cup (3, \infty)$.

Log: $x - 1 > 0$. Denominator: $\log(x - 1) \neq 0$.

$x > 1$. And $\log(x - 1) = 0 \iff x - 1 = 1 \iff x = 2$, so $x \neq 2$.

$x > 1$ and $x \neq 2$.

$(1, 2) \cup (2, \infty)$.

Each piece is a polynomial, so the only junction is $x = 2$.

$\lim_{x \to 2^-} f(x) = 4k - 1$.

$\lim_{x \to 2^+} f(x) = 6 + k = f(2)$.

$4k - 1 = 6 + k \Rightarrow 3k = 7 \Rightarrow k = \tfrac{7}{3}$.

Left: $\tfrac{28}{3} - 1 = \tfrac{25}{3}$. Right: $6 + \tfrac{7}{3} = \tfrac{25}{3}$. ✓

The junction is $x = 2$.

$\lim_{x \to 2^-} f(x) = 2a^2 - 2 = f(2)$.

$\lim_{x \to 2^+} f(x) = a + 4$.

$2a^2 - 2 = a + 4 \Rightarrow 2a^2 - a - 6 = 0 \Rightarrow (2a + 3)(a - 2) = 0 \Rightarrow a = 2$ or $a = -\tfrac{3}{2}$.

$a = 2$: both sides give $6$. $a = -\tfrac{3}{2}$: both sides give $\tfrac{5}{2}$. Both work.

Two junctions: $x = 1$ and $x = 3$.

At $x = 1$: $\lim_{x \to 1^-} f = a + 2$. At $x = 3$: $\lim_{x \to 3^-} f = 9 + b = f(3)$.

At $x = 1$: $\lim_{x \to 1^+} f = 1 + b = f(1)$. At $x = 3$: $\lim_{x \to 3^+} f = 18 - a$.

$a + 2 = 1 + b$ and $9 + b = 18 - a$, so $b = a + 1$ and $a + b = 9$. Then $2a + 1 = 9$: $a = 4$, $b = 5$.

At $1$: $4 + 2 = 6$ and $1 + 5 = 6$. At $3$: $9 + 5 = 14$ and $18 - 4 = 14$. ✓

The junction is $x = 4$, where the formula is replaced by the value $c$.

Same formula on both sides, and it gives $\tfrac{0}{0}$. Conjugate: $\dfrac{(x + 5) - 9}{(x - 4)(\sqrt{x + 5} + 3)} = \dfrac{1}{\sqrt{x + 5} + 3} \to \dfrac{1}{6}$.

Same formula, so the right-hand limit is also $\dfrac{1}{6}$.

$c = f(4) = \dfrac{1}{6}$.

$\dfrac{1}{\sqrt{9} + 3} = \dfrac{1}{6}$, and no division by zero remains. ✓

$x^2 - 9 = 0$ at $x = 3$ and $x = -3$.

$f(x) = \dfrac{(x - 3)(x + 2)}{(x - 3)(x + 3)} = \dfrac{x + 2}{x + 3}$ for $x \neq 3$. At $3$: both sides $\to \tfrac{5}{6}$. At $-3$: top $\to -1$, bottom $\to 0^{\mp}$, so $\lim_{x \to -3^-} f = +\infty$ and $\lim_{x \to -3^+} f = -\infty$.

$x = 3$: same finite limit, removable. $x = -3$: infinite limits, infinite.

Defining $f(3) = \tfrac{5}{6}$ removes the break at $x = 3$; nothing fixes $x = -3$.

The denominator is $0$ at $x = 2$, the same factor as inside $|\ |$.

$x > 2$: $|x - 2| = x - 2$, so $g = x \to 2$. $x < 2$: $|x - 2| = -(x - 2)$, so $g = -x \to -2$.

Two different finite limits, $-2$ and $2$: a jump.

Not removable, so there is no fixing value.

Breaks at $x = -2$ (a step), $x = 1$ (a hole) and $x = 3$ (an asymptote).

$x = -2$: left $2$, right $-1$. $x = 1$: both sides $2$, but $f(1) = -1$. $x = 3$: both sides $+\infty$.

$x = -2$: jump. $x = 1$: removable. $x = 3$: infinite.

Redefining $f(1) = 2$ fixes $x = 1$; the other two cannot be fixed.

$x \geq 0$ owns $0$: $f(0) = 0 + 2 = 2$.

$\lim_{x \to 0^-} \dfrac{\sin(2x)}{x} = \lim_{x \to 0^-} 2 \cdot \dfrac{\sin(2x)}{2x} = 2$. $\lim_{x \to 0^+} (x + 2) = 2$.

Both one-sided limits are $2$, so $\lim_{x \to 0} f(x) = 2$ exists.

$\lim_{x \to 0} f(x) = 2 = f(0)$.

All three conditions hold, so $f$ is continuous at $x = 0$.

$f(4) = 5$ (defined).

For $x \neq 4$: $\dfrac{(x - 4)(x + 2)}{x - 4} = x + 2 \to 6$ from both sides.

Both one-sided limits are $6$, so the limit exists and equals $6$.

$\lim_{x \to 4} f(x) = 6 \neq 5 = f(4)$.

The third condition fails, so $f$ is not continuous at $x = 4$ (a removable discontinuity).

$g(0) = 3 - 0 = 3$ (defined).

$\lim_{x \to 0^-} (e^x + 1) = 2$ and $\lim_{x \to 0^+} (3 - x) = 3$.

$2 \neq 3$, so $\lim_{x \to 0} g(x)$ does not exist.

There is no limit to compare with $g(0)$.

The second condition fails, so $g$ is not continuous at $x = 0$ (a jump discontinuity).