What is `lim_(xto0) (5^(x)-1)/(x)` equal to ?

Options

A. `log_(e)5`
B. `log_(5)e`
C. 5
D. 1

Explanation

To solve the limit \( \lim_{x \to 0} \frac{5^x - 1}{x} \), we can use a known limit property involving exponential functions.

### Step-by-Step Solution:

1. **Identify the limit form**: We are given the limit as \( x \) approaches 0 of the expression \( \frac{5^x - 1}{x} \). This is an indeterminate form \( \frac{0}{0} \) since \( 5^0 - 1 = 0 \) and \( x \to 0 \).

2. **Use the known limit property**: There is a standard result in calculus that states:
\[ \lim_{x \to 0} \frac{a^x - 1}{x} = \ln(a) \]
for any positive constant \( a \). In our case, \( a = 5 \).

3. **Apply the property**: Using the property mentioned, we can directly substitute \( a = 5 \):
\[ \lim_{x \to 0} \frac{5^x - 1}{x} = \ln(5) \]

4. **Conclusion**: Therefore, the limit is:
\[ \lim_{x \to 0} \frac{5^x - 1}{x} = \ln(5) \]

### Final Answer:
\[ \lim_{x \to 0} \frac{5^x - 1}{x} = \ln(5) \]