Evaluate the following limits :
`lim_(x to 2)(3^(x)+3^(3-x)-12)/(3^(3-x)-3^(x/2))`.

Answer

To evaluate the limit

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

we will follow these steps:

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

First, we substitute \( x = 2 \) into the limit expression to check if it results in an indeterminate form.

**Calculation:**
- Numerator:
\[ 3^2 + 3^{3-2} - 12 = 9 + 3 - 12 = 0. \]
- Denominator:
\[ 3^{3-2} - 3^{2/2} = 3 - 3 = 0. \]

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

**Hint:** Always check the limit by substituting the value first to see if it results in an 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 if \( \lim_{x \to c} \frac{f(x)}{g(x)} = \frac{0}{0} \), then:

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

provided the limit on the right exists.

**Hint:** Remember that L'Hôpital's Rule can only be applied to \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) forms.

### Step 3: Differentiate the Numerator and Denominator

Now we differentiate the numerator and denominator with respect to \( x \).

**Numerator:**
\[ f(x) = 3^x + 3^{3-x} - 12. \]
Using the chain rule:
\[ f'(x) = 3^x \ln(3) - 3^{3-x} \ln(3). \]

**Denominator:**
\[ g(x) = 3^{3-x} - 3^{x/2}. \]
Using the chain rule:
\[ g'(x) = -3^{3-x} \ln(3) + \frac{1}{2} 3^{x/2} \ln(3). \]

**Hint:** When differentiating exponential functions, remember to apply the chain rule and include the logarithm of the base.

### Step 4: Substitute \( x = 2 \) Again

Now we substitute \( x = 2 \) into the derivatives:

**Numerator:**
\[ f'(2) = 3^2 \ln(3) - 3^{3-2} \ln(3) = 9 \ln(3) - 3 \ln(3) = 6 \ln(3). \]

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

### Step 5: Calculate the Limit

Now we can compute the limit:

\[ \lim_{x \to 2} \frac{f'(x)}{g'(x)} = \frac{6 \ln(3)}{-\frac{3}{2} \ln(3)} = \frac{6}{-\frac{3}{2}} = 6 \cdot -\frac{2}{3} = -4. \]

### Final Answer

Thus, the limit evaluates to:

\[ \boxed{-4}. \]