The function provided is $f(x) = \frac{x^2 + 4}{x - 2}$, and we are tasked with determining whether it is continuous at $x = 3$.

To check continuity at a given point, $x = c$, the following conditions must be met:

1. $f(c)$ is defined.
2. The limit $\lim_{x \to c} f(x)$ exists.
3. The limit equals the function value: $\lim_{x \to c} f(x) = f(c)$.

Step 1: Check if $f(3)$ is defined

Substitute $x = 3$ into the function: $f(3) = \frac{3^2 + 4}{3 - 2} = \frac{9 + 4}{1} = \frac{13}{1} = 13$ So, $f(3)$ is defined and equals 13.

Step 2: Check if $\lim_{x \to 3} f(x)$ exists

To compute the limit, simply substitute $x = 3$ into the expression: $\lim_{x \to 3} f(x) = \lim_{x \to 3} \frac{x^2 + 4}{x - 2} = \frac{3^2 + 4}{3 - 2} = \frac{9 + 4}{1} = 13$ The limit exists and equals 13.

Step 3: Check if the limit equals $f(3)$

Since both the limit and the function value at $x = 3$ are 13, the third condition for continuity is satisfied.

Conclusion:

The function $f(x)$ is continuous at $x = 3$.

Would you like more details or have any questions?

Here are 5 related questions:

1. What happens if $f(x)$ is evaluated at $x = 2$?
2. How do we handle discontinuities in rational functions?
3. Can a function have a limit at a point but not be continuous at that point?
4. What are different types of discontinuities?
5. How can we determine if a piecewise function is continuous at a given point?

Tip: Always check for undefined points in rational functions, as they often cause discontinuities.