Determine the continuity of the function f(x) at x=3, where f(x) = (x3 − 27)/(x2 - 9) for x != 3 and f(x) = 5 for x = 3.

From the definition of $f(x)$, when $x=3$, $f(x)=5$.
So, $f(3) = 5$. The function is defined at $x=3$.

For $x \neq 3$, $f(x) = \frac{x^3 - 27}{x^2 - 9}$.
We can factor the numerator and denominator.
The numerator is a difference of cubes: $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$.
So, $x^3 - 27 = x^3 - 3^3 = (x-3)(x^2 + 3x + 9)$.
The denominator is a difference of squares: $a^2 - b^2 = (a-b)(a+b)$.
So, $x^2 - 9 = x^2 - 3^2 = (x-3)(x+3)$.

Now, substitute these factored forms into the limit expression:

Since $x \to 3$, $x \neq 3$, so $x-3 \neq 0$. We can cancel the $(x-3)$ terms.

Now, substitute $x=3$ into the simplified expression:

Simplify the fraction:

So, $\lim_{x \to 3} f(x) = \frac{9}{2}$.

We found $f(3) = 5$ and $\lim_{x \to 3} f(x) = \frac{9}{2}$.
Since $5 \neq \frac{9}{2}$, the third condition for continuity is not met.

Therefore, the function $f(x)$ is not continuous at $x=3$.

The function $f(x)$ is not continuous at $x=3$ because $\lim_{x \to 3} f(x) \neq f(3)$.