Math  /  Calculus

Question2.) (5pts) Evaluate the limit, if it exists. limx2x+2x3+8\lim _{x \rightarrow-2} \frac{x+2}{x^{3}+8}

Studdy Solution

STEP 1

1. We are given the function x+2x3+8 \frac{x+2}{x^3+8} .
2. We need to evaluate the limit as x x approaches 2-2.

STEP 2

1. Identify if direct substitution in the limit results in an indeterminate form.
2. Factor the denominator to simplify the expression.
3. Simplify the expression by canceling common factors.
4. Evaluate the limit after simplification.

STEP 3

Substitute x=2 x = -2 directly into the expression:
2+2(2)3+8=08+8=00 \frac{-2+2}{(-2)^3+8} = \frac{0}{-8+8} = \frac{0}{0}
This results in an indeterminate form 00 \frac{0}{0} .

STEP 4

Factor the denominator x3+8 x^3 + 8 using the sum of cubes formula:
x3+8=(x+2)(x22x+4) x^3 + 8 = (x + 2)(x^2 - 2x + 4)

STEP 5

Rewrite the original expression and simplify by canceling the common factor x+2 x + 2 :
x+2x3+8=x+2(x+2)(x22x+4) \frac{x+2}{x^3+8} = \frac{x+2}{(x+2)(x^2-2x+4)}
Cancel the common factor x+2 x + 2 :
=1x22x+4 = \frac{1}{x^2-2x+4}

STEP 6

Evaluate the limit of the simplified expression as x2 x \to -2 :
limx21x22x+4 \lim_{x \to -2} \frac{1}{x^2-2x+4}
Substitute x=2 x = -2 :
=1(2)22(2)+4 = \frac{1}{(-2)^2 - 2(-2) + 4} =14+4+4 = \frac{1}{4 + 4 + 4} =112 = \frac{1}{12}
The value of the limit is:
112 \boxed{\frac{1}{12}}

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