Evaluate the following limits :
`Lim_(x to a) (a^(x)-1)/(b^(x)-1)`

Answer

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

### Step 1: Rewrite the Limit
We start with the limit:
\[ L = \lim_{x \to a} \frac{a^x - 1}{b^x - 1} \]

### Step 2: Apply L'Hôpital's Rule
As \( x \to a \), both the numerator \( a^x - 1 \) and the denominator \( b^x - 1 \) approach 0. Hence, we can apply L'Hôpital's Rule, which states that if the limit results in an indeterminate form \( \frac{0}{0} \), we can take the derivative of the numerator and the derivative of the denominator.

### Step 3: Differentiate the Numerator and Denominator
The derivative of the numerator \( a^x - 1 \) is:
\[ \frac{d}{dx}(a^x) = a^x \ln(a) \]
The derivative of the denominator \( b^x - 1 \) is:
\[ \frac{d}{dx}(b^x) = b^x \ln(b) \]

### Step 4: Apply L'Hôpital's Rule
Now we can rewrite the limit using the derivatives:
\[ L = \lim_{x \to a} \frac{a^x \ln(a)}{b^x \ln(b)} \]

### Step 5: Substitute \( x = a \)
Now substitute \( x = a \):
\[ L = \frac{a^a \ln(a)}{b^a \ln(b)} \]

### Step 6: Simplify the Expression
Since \( a^a \) and \( b^a \) are constants, we can simplify the expression:
\[ L = \frac{\ln(a)}{\ln(b)} \cdot \frac{a^a}{b^a} \]

### Final Result
Thus, the limit evaluates to:
\[ L = \frac{\ln(a)}{\ln(b)} \]