`lim_(x to 0)(a^(x)-1)/x` is equal to :

Options

A. log a
B. log x
C. `log""1/a`
D. None of these

Answer

To solve the limit \( \lim_{x \to 0} \frac{a^x - 1}{x} \), we can follow these steps:

### Step 1: Identify the Form
First, we substitute \( x = 0 \) into the expression:
\[ \frac{a^0 - 1}{0} = \frac{1 - 1}{0} = \frac{0}{0} \]
This is an indeterminate form (0/0), so we can use L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, we differentiate the numerator and the denominator separately:
- The derivative of the numerator \( a^x - 1 \) is \( a^x \ln(a) \).
- The derivative of the denominator \( x \) is \( 1 \).

Thus, we can rewrite the limit as:
\[ \lim_{x \to 0} \frac{a^x \ln(a)}{1} \]

### Step 3: Evaluate the Limit
Now we can substitute \( x = 0 \) into the new expression:
\[ \lim_{x \to 0} a^x \ln(a) = a^0 \ln(a) = 1 \cdot \ln(a) = \ln(a) \]

### Final Answer
Therefore, the limit is:
\[ \lim_{x \to 0} \frac{a^x - 1}{x} = \ln(a) \]

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