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

Answer

To evaluate the limit

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

we will follow these steps:

### Step 1: Substitute \( x = 2 \)

First, we substitute \( x = 2 \) into the expression:

\[ 3^2 + 3^{2-1} - 12 = 9 + 3 - 12 = 0, \]
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 a \( \frac{0}{0} \) indeterminate form.

### Step 2: Apply L'Hôpital's Rule

Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that:

\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]

if the limit on the right side exists.

### Step 3: Differentiate the Numerator and Denominator

**Numerator:**
\[ f(x) = 3^x + 3^{x-1} - 12. \]
Differentiating \( f(x) \):
\[ f'(x) = 3^x \ln(3) + 3^{x-1} \ln(3). \]

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

### Step 4: Rewrite the Limit

Now we rewrite the limit using the derivatives:

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

### Step 5: Substitute \( x = 2 \) Again

Substituting \( x = 2 \):

**Numerator:**
\[ f'(2) = 3^2 \ln(3) + 3^{2-1} \ln(3) = 9 \ln(3) + 3 \ln(3) = 12 \ln(3). \]

**Denominator:**
\[ g'(2) = -\frac{1}{2} 3^{-1} \ln(3) + 3^{1-2} \ln(3) = -\frac{1}{2} \cdot \frac{1}{3} \ln(3) + \frac{1}{3} \ln(3) = -\frac{1}{6} \ln(3) + \frac{1}{3} \ln(3). \]
Simplifying the denominator:
\[ g'(2) = \frac{1}{3} \ln(3) - \frac{1}{6} \ln(3) = \frac{2}{6} \ln(3) - \frac{1}{6} \ln(3) = \frac{1}{6} \ln(3). \]

### Step 6: Final Limit Calculation

Now we have:

\[ \lim_{x \to 2} \frac{12 \ln(3)}{\frac{1}{6} \ln(3)} = 12 \cdot 6 = 72. \]

Thus, the value of the limit is

\[ \boxed{72}. \]