Consider the function `f(x)=(a^([x]+x)-1)/([x]+x)` where [.] denotes the greatest integer function.
What is `lim_(xto0^(+))f(x)` equal to?

Options

A. 1
B. In a
C. `1-a^(-1)`
D. Limit does not exist

Explanation

To find the limit \( \lim_{x \to 0^+} f(x) \) where \( f(x) = \frac{a^{[\cdot] + x} - 1}{[\cdot] + x} \) and \( [\cdot] \) denotes the greatest integer function, we will follow these steps:

### Step 1: Understand the greatest integer function
As \( x \) approaches \( 0 \) from the right (i.e., \( x \to 0^+ \)), the greatest integer function \( [x] \) will be \( 0 \). Therefore, we can rewrite the function as:
\[ f(x) = \frac{a^{[x] + x} - 1}{[x] + x} = \frac{a^{0 + x} - 1}{0 + x} = \frac{a^x - 1}{x} \]

### Step 2: Evaluate the limit
Now we need to evaluate the limit:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{a^x - 1}{x} \]

### Step 3: Apply L'Hôpital's Rule
Since both the numerator and denominator approach \( 0 \) as \( x \to 0^+ \), we can apply L'Hôpital's Rule:
\[ \lim_{x \to 0^+} \frac{a^x - 1}{x} = \lim_{x \to 0^+} \frac{d}{dx}(a^x - 1) / \frac{d}{dx}(x) \]
Calculating the derivatives:
- The derivative of \( a^x \) is \( a^x \ln(a) \).
- The derivative of \( x \) is \( 1 \).

Thus, we have:
\[ \lim_{x \to 0^+} \frac{a^x \ln(a)}{1} = a^0 \ln(a) = \ln(a) \]

### Step 4: Conclusion
Therefore, the limit is:
\[ \lim_{x \to 0^+} f(x) = \ln(a) \]

### Final Answer
\[ \lim_{x \to 0^+} f(x) = \ln(a) \]

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