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: Substitute \(x=3\) directly.
\[ \frac{3^3-3(3^2)}{3^2-5(3)+6} = \frac{27-27}{9-15+6} = \frac{0}{0} \] We get an indeterminate form.

Step 2: Factorize the numerator.
\[ x^3-3x^2 = x^2(x-3) \]

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

Step 4: Rewrite the limit expression.
\[ \lim_{x\to 3} \frac{x^2(x-3)}{(x-3)(x-2)} \]

Step 5: Cancel the common factor.
Since \((x-3)\) appears in numerator and denominator,
\[ \lim_{x\to 3} \frac{x^2}{x-2} \]

Step 6: Substitute \(x=3\).
\[ \frac{3^2}{3-2} = \frac{9}{1} = 9 \]

Step 7: Final conclusion.
Therefore, the value of the limit is
\[ \boxed{9} \]