Evaluate the following limits: `lim_(xrarr-2)((x^(3)+8)/(x+2))`

Options

A. 11
B. 12
C. 13
D. 14

Answer

To evaluate the limit \( \lim_{x \to -2} \frac{x^3 + 8}{x + 2} \), we can follow these steps:

### Step 1: Identify the expression
We start with the limit expression:
\[ \lim_{x \to -2} \frac{x^3 + 8}{x + 2} \]

### Step 2: Factor the numerator
Recognizing that \( x^3 + 8 \) is a sum of cubes, we can use the identity for factoring sums of cubes:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
In our case, \( a = x \) and \( b = 2 \). Thus, we can rewrite:
\[ x^3 + 8 = x^3 + 2^3 = (x + 2)(x^2 - 2x + 4) \]

### Step 3: Substitute the factorization into the limit
Substituting the factorization back into the limit gives:
\[ \lim_{x \to -2} \frac{(x + 2)(x^2 - 2x + 4)}{x + 2} \]

### Step 4: Cancel common factors
Since \( x + 2 \) appears in both the numerator and the denominator, we can cancel it (as long as \( x \neq -2 \)):
\[ \lim_{x \to -2} (x^2 - 2x + 4) \]

### Step 5: Evaluate the limit
Now we can directly substitute \( x = -2 \) into the remaining expression:
\[ = (-2)^2 - 2(-2) + 4 \]
Calculating this gives:
\[ = 4 + 4 + 4 = 12 \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to -2} \frac{x^3 + 8}{x + 2} = 12 \]

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