Evaluate `lim_(x rarr 2) (3^(x)+3^(3 - x)-12)/(3^((-x)/(2))-3^(1-x))`

Answer

To evaluate the limit

\[ \lim_{x \to 2} \frac{3^x + 3^{3 - x} - 12}{3^{-\frac{x}{2}} - 3^{1 - x}}, \]

we will 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:

\[ 3^2 + 3^{3 - 2} - 12 = 9 + 3 - 12 = 0, \]
and
\[ 3^{-\frac{2}{2}} - 3^{1 - 2} = 3^{-1} - 3^{-1} = \frac{1}{3} - \frac{1}{3} = 0. \]

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

**Hint:** Check if substituting the limit value results in an indeterminate form.

### Step 2: Apply L'Hôpital's Rule
Since we have \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the derivative of the denominator.

**Numerator:**
\[ \frac{d}{dx}(3^x + 3^{3 - x} - 12) = 3^x \ln(3) - 3^{3 - x} \ln(3). \]

**Denominator:**
\[ \frac{d}{dx}(3^{-\frac{x}{2}} - 3^{1 - x}) = -\frac{1}{2} 3^{-\frac{x}{2}} \ln(3) + 3^{1 - x} \ln(3). \]

### Step 3: Rewrite the limit using derivatives
Now we can rewrite the limit as:

\[ \lim_{x \to 2} \frac{3^x \ln(3) - 3^{3 - x} \ln(3)}{-\frac{1}{2} 3^{-\frac{x}{2}} \ln(3) + 3^{1 - x} \ln(3)}. \]

### Step 4: Factor out \( \ln(3) \)
Factoring out \( \ln(3) \) from both the numerator and denominator gives:

\[ \lim_{x \to 2} \frac{\ln(3) (3^x - 3^{3 - x})}{\ln(3) \left(-\frac{1}{2} 3^{-\frac{x}{2}} + 3^{1 - x}\right)}. \]

This simplifies to:

\[ \lim_{x \to 2} \frac{3^x - 3^{3 - x}}{-\frac{1}{2} 3^{-\frac{x}{2}} + 3^{1 - x}}. \]

### Step 5: Substitute \( x = 2 \) again
Now substituting \( x = 2 \):

Numerator:
\[ 3^2 - 3^{3 - 2} = 9 - 3 = 6. \]

Denominator:
\[ -\frac{1}{2} 3^{-1} + 3^{1 - 2} = -\frac{1}{2} \cdot \frac{1}{3} + \frac{1}{3} = -\frac{1}{6} + \frac{1}{3} = -\frac{1}{6} + \frac{2}{6} = \frac{1}{6}. \]

### Step 6: Final calculation
Now, we can compute the limit:

\[ \lim_{x \to 2} \frac{6}{\frac{1}{6}} = 6 \cdot 6 = 36. \]

Thus, the final answer is

\[ \boxed{36}. \]