Evaluate : `lim_(x to 0)(2^(2+x)-9)/x`.

Answer

To evaluate the limit \( \lim_{x \to 0} \frac{2^{2+x} - 9}{x} \), we can follow these steps:

### Step 1: Substitute \( x = 0 \)

First, we substitute \( x = 0 \) into the expression:

\[ 2^{2+x} = 2^{2+0} = 2^2 = 4 \]

So, the expression becomes:

\[ \frac{2^{2+0} - 9}{0} = \frac{4 - 9}{0} = \frac{-5}{0} \]

### Step 2: Analyze the result

Since we are dividing by zero, we need to analyze the limit further. The expression \( \frac{-5}{0} \) indicates that the limit approaches either \( +\infty \) or \( -\infty \) depending on the direction from which \( x \) approaches 0.

### Step 3: Use L'Hôpital's Rule

Since we have an indeterminate form of type \( \frac{0}{0} \), we can apply L'Hôpital's Rule. This rule states that if the limit results in \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), we can take the derivative of the numerator and the derivative of the denominator.

#### Derivative of the numerator:

The numerator is \( 2^{2+x} - 9 \). The derivative of \( 2^{2+x} \) with respect to \( x \) is:

\[ \frac{d}{dx}(2^{2+x}) = 2^{2+x} \ln(2) \]

So, the derivative of the numerator is:

\[ \frac{d}{dx}(2^{2+x} - 9) = 2^{2+x} \ln(2) \]

#### Derivative of the denominator:

The denominator is simply \( x \), and its derivative is:

\[ \frac{d}{dx}(x) = 1 \]

### Step 4: Apply L'Hôpital's Rule

Now we can apply L'Hôpital's Rule:

\[ \lim_{x \to 0} \frac{2^{2+x} - 9}{x} = \lim_{x \to 0} \frac{2^{2+x} \ln(2)}{1} \]

### Step 5: Substitute \( x = 0 \) again

Now substitute \( x = 0 \):

\[ 2^{2+0} \ln(2) = 4 \ln(2) \]

### Final Result

Thus, the limit is:

\[ \lim_{x \to 0} \frac{2^{2+x} - 9}{x} = 4 \ln(2) \]

### Summary

The final answer is:

\[ \lim_{x \to 0} \frac{2^{2+x} - 9}{x} = 4 \ln(2) \]