1. Find the discontinuity points of the function $f(x)$.\nThe problem states that the function is discontinuous at $x=2$ and $x=3$ because the values of $f(x)$ near these points do not match the function values themselves. The points are specifically $(2,??)$ and $(3,\theta)$. These are jump or undefined value discontinuities observable from the graph.\n\n2. Find the domain of $f(x) = \sqrt{2-x} + \sqrt{3+x}$.\nSince both expressions inside the square roots must be non-negative, we solve the inequalities:\n\n$$2 - x \geq 0 \implies x \leq 2$$\n$$3 + x \geq 0 \implies x \geq -3$$\n\nHence, the domain of $f(x)$ is all $x$ such that $$-3 \leq x \leq 2.$$\n\n3. Find $\dfrac{d}{dx} [\log_a e]$.\nSince $\log_a e$ is the logarithm of the constant $e$ with base $a$, and is thus a constant with respect to $x$, the derivative is:\n\n$$\frac{d}{dx} [\log_a e] = 0.$$\n\n4. Find the limit $$\lim_{x \to 0} \frac{\sqrt{2} + \sec x}{\cos (\pi - \tan x)}.$$\nStep 1: Evaluate inner functions at $x=0$. Recall that $\sec 0 = 1$ and $\tan 0 = 0$.\nStep 2: Calculate numerator at $x=0$:\n$$\sqrt{2} + \sec 0 = \sqrt{2} + 1.$$\nStep 3: Simplify denominator at $x=0$:\n$$\cos (\pi - \tan 0) = \cos(\pi - 0) = \cos \pi = -1.$$\nStep 4: Substitute back into the limit expression:\n$$\lim_{x \to 0} \frac{\sqrt{2} + \sec x}{\cos (\pi - \tan x)} = \frac{\sqrt{2} + 1}{-1} = - (\sqrt{2} + 1).$$\n\nFinal answers:\n1. Discontinuities at $x=2$ and $x=3$.\n2. Domain: $[-3, 2]$.\n3. Derivative: $0$.\n4. Limit: $- (\sqrt{2} + 1)$.