`lim_(x to 0) (a^(x)-b^(x))/x=`

Options

A. log (ab)
B. `log a/b`
C. `log b/a`
D. none

Answer

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

### Step 1: Identify the form of the limit
When we substitute \( x = 0 \) directly into the expression, we get:
\[ \frac{a^0 - b^0}{0} = \frac{1 - 1}{0} = \frac{0}{0} \]
This is an indeterminate form, so we can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
L'Hôpital's Rule states that if we have an indeterminate form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), we can take the derivative of the numerator and the derivative of the denominator:
\[ \lim_{x \to 0} \frac{a^x - b^x}{x} = \lim_{x \to 0} \frac{(a^x \ln a) - (b^x \ln b)}{1} \]

### Step 3: Evaluate the new limit
Now we need to evaluate:
\[ \lim_{x \to 0} \left( a^x \ln a - b^x \ln b \right) \]
As \( x \to 0 \), \( a^x \to 1 \) and \( b^x \to 1 \). Thus, we have:
\[ \lim_{x \to 0} (a^x \ln a - b^x \ln b) = 1 \cdot \ln a - 1 \cdot \ln b = \ln a - \ln b \]

### Step 4: Simplify the expression
Using the properties of logarithms, we can rewrite:
\[ \ln a - \ln b = \ln \left( \frac{a}{b} \right) \]

### Final Result
Thus, the limit is:
\[ \lim_{x \to 0} \frac{a^x - b^x}{x} = \ln \left( \frac{a}{b} \right) \]