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. 0
B. In a
C. `1-a^(-1)`
D. Limit does not exist

Explanation

To find the limit of the function \( f(x) = \frac{a^{[\text{x}] + x} - 1}{[\text{x}] + x} \) as \( x \) approaches 0 from the left (denoted as \( \lim_{x \to 0^-} f(x) \)), we will follow these steps:

### Step 1: Understand the greatest integer function
The greatest integer function \( [x] \) gives the largest integer less than or equal to \( x \). As \( x \) approaches 0 from the left (i.e., \( x \to 0^- \)), \( [x] \) will be -1. Therefore, we can rewrite the function as:

\[ f(x) = \frac{a^{[-1] + x} - 1}{[-1] + x} = \frac{a^{-1 + x} - 1}{-1 + x} \]

### Step 2: Substitute \( x \) with \( 0 - h \)
Let \( x = 0 - h \) where \( h \) is a small positive number approaching 0. Then, we have:

\[ f(0 - h) = \frac{a^{-1 + (0 - h)} - 1}{-1 + (0 - h)} = \frac{a^{-1 - h} - 1}{-1 - h} \]

### Step 3: Simplify the expression
Now, we simplify the expression:

\[ f(0 - h) = \frac{a^{-1 - h} - 1}{-1 - h} \]

### Step 4: Apply the limit
We need to find the limit as \( h \to 0^+ \):

\[ \lim_{h \to 0^+} f(0 - h) = \lim_{h \to 0^+} \frac{a^{-1 - h} - 1}{-1 - h} \]

### Step 5: Use L'Hôpital's Rule
Since both the numerator and denominator approach 0 as \( h \to 0^+ \), we can apply L'Hôpital's Rule:

1. Differentiate the numerator:
\[ \frac{d}{dh}(a^{-1 - h} - 1) = -a^{-1 - h} \ln(a) \]

2. Differentiate the denominator:
\[ \frac{d}{dh}(-1 - h) = -1 \]

Now applying L'Hôpital's Rule:

\[ \lim_{h \to 0^+} \frac{-a^{-1 - h} \ln(a)}{-1} = \lim_{h \to 0^+} a^{-1 - h} \ln(a) \]

### Step 6: Evaluate the limit
As \( h \to 0 \), \( a^{-1 - h} \) approaches \( a^{-1} \):

\[ \lim_{h \to 0^+} a^{-1 - h} \ln(a) = a^{-1} \ln(a) \]

### Final Result
Thus, the limit is:

\[ \lim_{x \to 0^-} f(x) = a^{-1} \ln(a) \]