Skill: Evaluate limits algebraically

To evaluate a limit algebraically, first try direct substitution: replace the variable with the number it approaches.

• If substitution gives a real number, that is the limit.
• If substitution gives an indeterminate form such as $\frac{0}{0}$, simplify the expression algebraically before substituting again.

Example

Evaluate

$$\lim_{x\to 3}\frac{x^2-9}{x-3}.$$

Step 1: Try direct substitution.

Substitute $x=3$:

$$\frac{3^2-9}{3-3} = \frac{9-9}{0} = \frac{0}{0}.$$

Since $\frac{0}{0}$ is indeterminate, we need to simplify.

Step 2: Factor the numerator.

The numerator is a difference of squares:

$$x^2-9=(x-3)(x+3).$$

So the expression becomes

$$\frac{(x-3)(x+3)}{x-3}.$$

Step 3: Cancel the common factor.

For values of $x$ near $3$ but not equal to $3$, we can cancel $x-3$:

$$\frac{(x-3)(x+3)}{x-3}=x+3.$$

Therefore,

$$\lim_{x\to 3}\frac{x^2-9}{x-3} = \lim_{x\to 3}(x+3).$$

Step 4: Substitute again.

$$3+3=6.$$

Thus,

$$\boxed{\lim_{x\to 3}\frac{x^2-9}{x-3}=6}.$$

The original function is undefined at $x=3$, but the limit is $6$ because the nearby values approach $6$. A limit describes what happens near the input, not necessarily the function’s value exactly at that input.