Question

The value of

Hint:

Whenever direct substitution gives \(\frac{0}{0}\), factorize numerator and denominator to simplify the expression before evaluating the limit.

Answer 1

Correct Answer: 9

Step 1: Set up the limit.
We want \[ \lim_{x\to 3} \frac{x^3 - 3x^2}{x^2 - 5x + 6} \] Putting $x=3$ gives $0/0$, so we factor.

Step 2: Factor the top.
\[ x^3 - 3x^2 = x^2(x-3) \]

Step 3: Factor the bottom.
\[ x^2 - 5x + 6 = (x-3)(x-2) \]

Step 4: Cancel and substitute.
The $(x-3)$ cancels, leaving $\dfrac{x^2}{x-2}$. At $x=3$ this is $\dfrac{9}{1} = 9$.

Step 5: Answer.
\[ \boxed{9} \]