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Evaluate zero-over-zero limits by factoring and rationalizing

Remove the zero factor before substituting

The form $0/0$ does not determine a limit. Try factoring polynomials, or multiply by a conjugate when square roots are involved. Cancel only where the canceled factor is nonzero; the simplified expression can still determine the limit nearby.

Problem

Evaluate Example 1. $x \to - 1 lim \frac{x ^{3} - 3 x - 2}{x ^{2} - x - 2}$ . Example 2. $x \to 3 lim \frac{x + 13 - 2 x + 1}{x ^{2} - 9}$ .

Answer

Example 1:

0

. Example 2:

- 1/16

The original expressions are undefined at the approach points, but their limits exist.

Step-by-step solution

1. Factor the polynomial ratio

Direct substitution at $x = - 1$ gives $0/0$ . Factor:

x^{3} - 3 x - 2 = (x + 1)^{2} (x - 2), x^{2} - x - 2 = (x + 1) (x - 2) .

For nearby $x \neq = - 1, 2$ , the ratio equals $x + 1$ . Thus its limit at $- 1$ is 0.

2. Use the conjugate for the square roots

At $x = 3$ , the second expression also gives $0/0$ . Multiply numerator and denominator by $x + 13 + 2 x + 1$ . The numerator becomes

(x + 13) - 4 (x + 1) = - 3 (x - 3) .

Since $x^{2} - 9 = (x - 3) (x + 3)$ , cancel $x - 3$ for nearby $x \neq = 3$ .

3. Substitute into the simplified expression

The limit is

x \to 3 lim \frac{- 3}{( x + 3 ) ( x + 13 + 2 x + 1 )} = \frac{- 3}{6 ( 4 + 4 )} = - \frac{1}{16} .

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