The value of `lim_(x to 2) (3^(x//2)-3)/(x^(3)-9)` is

Options

A. `0`
B. `(1)/(3)`
C. `(1)/(6)`
D. ln 3

Explanation

To find the limit \( \lim_{x \to 2} \frac{3^{(x/2)} - 3}{x^3 - 9} \), we can follow these steps:

### Step 1: Substitute the limit value
First, we substitute \( x = 2 \) into the expression to check if we get an indeterminate form.

\[ \text{Numerator: } 3^{(2/2)} - 3 = 3^1 - 3 = 3 - 3 = 0 \]
\[ \text{Denominator: } 2^3 - 9 = 8 - 9 = -1 \]

Since the numerator is \( 0 \) and the denominator is \( -1 \), we do not have an indeterminate form. However, let's check the limit again to ensure we are not missing anything.

### Step 2: Factor the denominator
The denominator \( x^3 - 9 \) can be factored using the difference of cubes formula:

\[ x^3 - 9 = x^3 - 3^2 = (x - 2)(x^2 + 2x + 4) \]

### Step 3: Rewrite the limit
Now we can rewrite the limit:

\[ \lim_{x \to 2} \frac{3^{(x/2)} - 3}{(x - 2)(x^2 + 2x + 4)} \]

### Step 4: Factor the numerator
Next, we can factor the numerator \( 3^{(x/2)} - 3 \). We can use the fact that \( a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + \ldots + b^{n-1}) \). Here, we can use the derivative to find the limit as \( x \) approaches \( 2 \):

Using L'Hôpital's Rule since we have a \( \frac{0}{0} \) form:

\[ \text{Differentiate the numerator: } \frac{d}{dx}(3^{(x/2)} - 3) = \frac{1}{2} \cdot 3^{(x/2)} \ln(3) \]
\[ \text{Differentiate the denominator: } \frac{d}{dx}(x^3 - 9) = 3x^2 \]

### Step 5: Apply L'Hôpital's Rule
Now we can apply L'Hôpital's Rule:

\[ \lim_{x \to 2} \frac{\frac{1}{2} \cdot 3^{(x/2)} \ln(3)}{3x^2} \]

### Step 6: Substitute \( x = 2 \)
Now substitute \( x = 2 \):

\[ = \frac{\frac{1}{2} \cdot 3^{(2/2)} \ln(3)}{3 \cdot 2^2} = \frac{\frac{1}{2} \cdot 3^1 \ln(3)}{3 \cdot 4} = \frac{\frac{3}{2} \ln(3)}{12} = \frac{\ln(3)}{8} \]

### Final Answer
Thus, the value of the limit is:

\[ \lim_{x \to 2} \frac{3^{(x/2)} - 3}{x^3 - 9} = \frac{\ln(3)}{8} \]