Flow Control Valve Restriction (Civil & Environmental Engineering) Solution

Step-by-Step Solution

1. Direct substitution check: Plugging $x = 3$ directly into the velocity function:

   $$V(3) = \frac{3^2 - 5(3) + 6}{3 - 3} = \frac{9 - 15 + 6}{0} = \frac{0}{0}$$
   Direct evaluation yields the indeterminate form $\frac{0}{0}$. The function value is undefined at $x = 3$, meaning the raw formula experiences a removable singularity at that exact aperture width.

2. Algebraic factoring and limit evaluation: Factor the quadratic polynomial in the numerator by finding factors of $+6$ that sum to $-5$, which are $-2$ and $-3$:

   $$x^2 - 5x + 6 = (x - 2)(x - 3)$$

   Rewrite the limit expression:

   $$\lim_{x\to 3} V(x) = \lim_{x\to 3} \frac{(x - 2)(x - 3)}{x - 3}$$

   Since $x \to 3$ implies $x \neq 3$, cancel the common zero-inducing factor $(x - 3)$:

   $$\lim_{x\to 3} V(x) = \lim_{x\to 3} (x - 2)$$

   Apply direct substitution to the simplified expression:

   $$3 - 2 = 1\text{ m/s}$$

Conclusion: The limiting flow velocity as the aperture width approaches $3\text{ cm}$ is $1\text{ m/s}$. Factoring eliminates the mathematical singularity, allowing the civil/environmental engineer to determine the continuous flow rate through the valve without division-by-zero errors.