What is `lim_(xto2)(2-x)/(x^(3)-8)` equal to ?

Options

A. `(1)/(8)`
B. `-(1)/(8)`
C. `(1)/(12)`
D. `-(1)/(12)`

Answer

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

### Step 1: Substitute the limit value
First, we substitute \( x = 2 \) into the function to check the form of the limit:

\[ \text{Numerator: } 2 - 2 = 0 \]
\[ \text{Denominator: } 2^3 - 8 = 8 - 8 = 0 \]

Since both the numerator and denominator evaluate to 0, we have an indeterminate form \( \frac{0}{0} \).

**Hint:** When you encounter \( \frac{0}{0} \), consider simplifying the expression.

### Step 2: Factor the denominator
The denominator \( x^3 - 8 \) can be factored using the difference of cubes formula:

\[ x^3 - 8 = (x - 2)(x^2 + 2x + 4) \]

**Hint:** Remember the difference of cubes formula: \( a^3 - b^3 = (a-b)(a^2 + ab + b^2) \).

### Step 3: Rewrite the limit
Now we can rewrite the limit using the factored form of the denominator:

\[ \lim_{x \to 2} \frac{2 - x}{(x - 2)(x^2 + 2x + 4)} \]

Notice that \( 2 - x = -(x - 2) \). Thus, we can rewrite the limit as:

\[ \lim_{x \to 2} \frac{-(x - 2)}{(x - 2)(x^2 + 2x + 4)} \]

**Hint:** Look for opportunities to cancel common factors in the numerator and denominator.

### Step 4: Cancel common factors
We can cancel \( (x - 2) \) from the numerator and denominator:

\[ \lim_{x \to 2} \frac{-1}{x^2 + 2x + 4} \]

**Hint:** After canceling, ensure that the limit is still valid by checking the new expression.

### Step 5: Substitute again
Now we substitute \( x = 2 \) into the simplified expression:

\[ x^2 + 2x + 4 = 2^2 + 2 \cdot 2 + 4 = 4 + 4 + 4 = 12 \]

So the limit becomes:

\[ \lim_{x \to 2} \frac{-1}{12} = -\frac{1}{12} \]

### Final Answer
Thus, the limit is:

\[ \boxed{-\frac{1}{12}} \]

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