Evaluate: `lim_(xto0)(log(1-3x))/(5^(x)-1)`

Options

A. `(-3)/(log5)`
B. `(-1)/(log5)`
C. `3/(log5)`
D. `1/(log5)`

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{\log(1 - 3x)}{5^x - 1}, \]

we start by substituting \( x = 0 \) into the expression:

1. **Substituting \( x = 0 \)**:
\[ \log(1 - 3 \cdot 0) = \log(1) = 0, \]
and
\[ 5^0 - 1 = 1 - 1 = 0. \]
This gives us the indeterminate form \( \frac{0}{0} \).

2. **Applying L'Hôpital's Rule**:
Since we have the indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that if \( \lim_{x \to c} \frac{f(x)}{g(x)} = \frac{0}{0} \) or \( \frac{\pm \infty}{\pm \infty} \), then:
\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}, \]
provided the limit on the right exists.

3. **Finding the derivatives**:
- The derivative of the numerator \( f(x) = \log(1 - 3x) \):
\[ f'(x) = \frac{1}{1 - 3x} \cdot (-3) = \frac{-3}{1 - 3x}. \]
- The derivative of the denominator \( g(x) = 5^x - 1 \):
\[ g'(x) = 5^x \ln(5). \]

4. **Applying L'Hôpital's Rule**:
Now we can rewrite the limit:
\[ \lim_{x \to 0} \frac{\log(1 - 3x)}{5^x - 1} = \lim_{x \to 0} \frac{-3/(1 - 3x)}{5^x \ln(5)}. \]

5. **Substituting \( x = 0 \) again**:
Now we substitute \( x = 0 \) into the new limit:
\[ = \frac{-3/(1 - 3 \cdot 0)}{5^0 \ln(5)} = \frac{-3/1}{1 \cdot \ln(5)} = \frac{-3}{\ln(5)}. \]

Thus, the final answer is:

\[ \lim_{x \to 0} \frac{\log(1 - 3x)}{5^x - 1} = \frac{-3}{\ln(5)}. \]