`lim_(xrarr0) (a^x-b^x)/(e^x-1)` is equal to

Options

A. `log_(e)((a)/(b))`
B. `log_(e)((b)/(a))`
C. `log_e ab`
D. `log_e(a+b)`

Explanation

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

### Step 1: Rewrite the limit
We start with the limit expression:
\[ \lim_{x \to 0} \frac{a^x - b^x}{e^x - 1} \]
We can rewrite the numerator by adding and subtracting 1:
\[ = \lim_{x \to 0} \frac{(a^x - 1) - (b^x - 1)}{e^x - 1} \]

### Step 2: Factor the limit
Now, we can separate the limit:
\[ = \lim_{x \to 0} \frac{a^x - 1}{e^x - 1} - \lim_{x \to 0} \frac{b^x - 1}{e^x - 1} \]

### Step 3: Apply L'Hôpital's Rule
Both limits are of the form \( \frac{0}{0} \) as \( x \to 0 \). We can apply L'Hôpital's Rule:
\[ \lim_{x \to 0} \frac{a^x - 1}{e^x - 1} = \lim_{x \to 0} \frac{a^x \ln a}{e^x} = \frac{1 \cdot \ln a}{1} = \ln a \]
Similarly,
\[ \lim_{x \to 0} \frac{b^x - 1}{e^x - 1} = \lim_{x \to 0} \frac{b^x \ln b}{e^x} = \frac{1 \cdot \ln b}{1} = \ln b \]

### Step 4: Substitute the limits back
Now substituting back into our expression:
\[ = \ln a - \ln b \]

### Step 5: Use logarithmic properties
Using the property of logarithms:
\[ \ln a - \ln b = \ln \left(\frac{a}{b}\right) \]

### Step 6: Final result
Thus, we have:
\[ \lim_{x \to 0} \frac{a^x - b^x}{e^x - 1} = \ln \left(\frac{a}{b}\right) \]

### Conclusion
The final answer is:
\[ \lim_{x \to 0} \frac{a^x - b^x}{e^x - 1} = \ln \left(\frac{a}{b}\right) \]