Q.Evaluate $\lim_{x \to 3} \dfrac{x^2 - 9}{x - 3}$.

Concept: Limit of an indeterminate form $\frac{0}{0}$ — factor and cancel.

Direct substitution gives $\frac{3^2 - 9}{3 - 3} = \frac{0}{0}$, which is indeterminate. The numerator is a difference of squares:

Cancel the common factor $(x - 3)$ (valid since $x \to 3$ means $x \neq 3$):

Now substitute $x = 3$:

When you substitute $x = 3$ directly into $\frac{x^2 - 9}{x - 3}$, both numerator and denominator vanish: you get $\frac{0}{0}$, which is indeterminate. This signals that the function has a "hole" at $x = 3$ rather than being truly undefined everywhere near that point. The key insight is that $x^2 - 9$ is a difference of squares, so it factors as $(x - 3)(x + 3)$. Once we cancel the common factor $x - 3$ (valid for all $x \neq 3$), the simplified expression reveals what value the function approaches.

Step-by-step evaluation

1. Recognize the indeterminate form. Plugging in $x = 3$ gives

This tells us we cannot evaluate the limit by direct substitution; algebraic simplification is needed.

2. Factor the numerator. The numerator $x^2 - 9$ is a difference of squares:

So the original expression becomes

3. Cancel the common factor. For every $x \neq 3$, we can cancel $x - 3$ from numerator and denominator:

The limit asks what happens as $x$ approaches 3, not at $x = 3$, so this cancellation is legitimate.

4. Evaluate the simplified limit. Now