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 first check the form of the limit by substituting \( x = 0 \):

- The numerator becomes \( \log(1 - 3 \cdot 0) = \log(1) = 0 \).
- The denominator becomes \( 5^0 - 1 = 1 - 1 = 0 \).

Since we have a \( \frac{0}{0} \) indeterminate form, we can apply L'Hôpital's Rule, which states that if the limit results in an indeterminate form, we can take the derivative of the numerator and the derivative of the denominator separately.

### Step 1: Differentiate the numerator and denominator

1. **Numerator**:
\[ \frac{d}{dx}[\log(1 - 3x)] = \frac{-3}{1 - 3x} \]
(using the chain rule).

2. **Denominator**:
\[ \frac{d}{dx}[5^x - 1] = 5^x \log(5) \]
(using the derivative of \( a^x \) which is \( a^x \log(a) \)).

### Step 2: Apply L'Hôpital's Rule

Now we can rewrite the limit using the derivatives we found:

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

### Step 3: Substitute \( x = 0 \)

Now we substitute \( x = 0 \) into the limit:

- The numerator becomes:
\[ -\frac{3}{1 - 3 \cdot 0} = -\frac{3}{1} = -3. \]

- The denominator becomes:
\[ 5^0 \log(5) = 1 \cdot \log(5) = \log(5). \]

### Step 4: Final Calculation

Thus, we have:

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

### Conclusion

The final answer is:

\[ -\frac{3}{\log(5)}. \]