How to Evaluate a Limit by Factoring the Numerator

Step-by-step solution

1. Try direct substitution.

   $$\frac{(-2)^2 + 5(-2) + 6}{-2 + 2} = \frac{0}{0}$$

   Why: Zero over zero means the limit needs work before substituting.

2. Factor the numerator.

   $$x^2 + 5x + 6 = (x + 2)(x + 3)$$

   Why: The factor (x + 2) is what creates the zero on the bottom.

3. Cancel the common factor.

   $$\frac{(x + 2)(x + 3)}{x + 2} = x + 3 \quad (x \neq -2)$$

   Why: The cancelled expression matches the original everywhere except at x = -2.

4. Substitute into the simpler expression.

   $$-2 + 3 = 1$$

   Why: The simplified expression is continuous at -2, so substitution now works.

Practice problems

Use the same method. Work each problem on paper, then open the answer to check.

1. 1

   $\lim_{x \to -3} \frac{x^2 + 7x + 12}{x + 3}$

   Show answer

   $1$

2. 2

   $\lim_{x \to 1} \frac{x^2 + 3x - 4}{x - 1}$

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   $5$

3. 3

   $\lim_{x \to -1} \frac{x^2 + 4x + 3}{x + 1}$

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   $2$

4. 4

   $\lim_{x \to 2} \frac{x^2 + x - 6}{x - 2}$

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   $5$

5. 5

   $\lim_{x \to -4} \frac{x^2 + 9x + 20}{x + 4}$

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   $1$