Evaluating the limit as $x \to +\infty$:
Factor $x^2$ out of the square root in the denominator (for $x \gt 0$, $\sqrt{x^2} = x$):

Calculations for profit maximization:

• Revenue function: $R(x) = 200x$
• Cost function: $C(x) = 500,000 + 80x + 0.003x^2$
• Profit function $P(x) = R(x) - C(x)$:
  $$P(x) = 200x - (500,000 + 80x + 0.003x^2) = -500,000 + 120x - 0.003x^2$$
• Differentiate with respect to $x$ and set $P'(x) = 0$:
  $$P'(x) = 120 - 0.006x = 0$$
  $$0.006x = 120 \implies x = \frac{120}{0.006} = 20,000$$
• Second derivative test: $P''(x) = -0.006 \lt 0$, confirming that $x = 20,000$ yields a maximum.
  Since $20,000 \le 30,000$ (within the production capacity), 20,000 units maximize profit. Option 2 is correct.

According to Clairaut's (or Schwarz's) Theorem on mixed partial derivatives:
If a function $z = f(x, y)$ and its partial derivatives $f_x, f_y, f_{xy}, f_{yx}$ are continuous in an open region, then the mixed second-order partial derivatives are identical:

Since $z = 3x^2 y - x \sin xy$ is composed of polynomial and trigonometric functions, it is infinitely differentiable everywhere on $\mathbb{R}^2$.