Examples of calculating limits by simplifying fractions

Example 1:
$\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$
Step 1: Plug in the limit directly
$\frac{2^2 - 4}{2 - 2} = \frac{4 - 4}{0} = \frac{0}{0}$

This is an indeterminate form, so we need to simplify.

Step 2: Factor and simplify
$x^2 - 4 = (x - 2)(x + 2)$

So the expression becomes:

$\frac{(x - 2)(x + 2)}{x - 2}$

Now cancel the common factor:

$= x + 2$

Step 3: Take the limit of the simplified expression
$\lim_{x \to 2} x + 2 = 2 + 2 = \boxed{4}$

Example 2:
$\lim_{x \to 1} \frac{x^3 - 1}{x - 1}$
Step 1: Direct substitution

$\frac{1^3 - 1}{1 - 1} = \frac{0}{0}$

This is indeterminate, so we simplify.

Step 2: Use a factoring identity

This is a difference of cubes:

$x^3 - 1 = (x - 1)(x^2 + x + 1)$

So the expression becomes:

$\frac{(x - 1)(x^2 + x + 1)}{x - 1}$

Step 3: Cancel common factor
$\frac{(x - 1)(x^2 + x + 1)}{x - 1} = x^2 + x + 1 \quad \text{(for } x \ne 1 \text{)}$

Step 4: Take the limit
$\lim_{x \to 1} x^2 + x + 1 = 1^2 + 1 + 1 = \boxed{3}$

Final Answer:
$\lim_{x \to 1} \frac{x^3 - 1}{x - 1} = \boxed{3}$

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