`lim_(x to 0) (2^(x)-1)/(x) +lim_(x to 0) (3^(x)-1)/(x) - lim_(x to 0) ((6^(x)-1)/(x))` equals :

Options

A. 1
B. 2
C. 0
D. 4

Explanation

To solve the limit expression

\[ \lim_{x \to 0} \frac{2^x - 1}{x} + \lim_{x \to 0} \frac{3^x - 1}{x} - \lim_{x \to 0} \frac{6^x - 1}{x}, \]

we can use the property of limits for exponential functions. The property states that:

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

where \(\log\) is the natural logarithm (base \(e\)).

### Step 1: Apply the limit property to each term

1. **First Limit:**
\[ \lim_{x \to 0} \frac{2^x - 1}{x} = \log(2) \]

2. **Second Limit:**
\[ \lim_{x \to 0} \frac{3^x - 1}{x} = \log(3) \]

3. **Third Limit:**
\[ \lim_{x \to 0} \frac{6^x - 1}{x} = \log(6) \]

### Step 2: Substitute the limits back into the expression

Now we can substitute these results back into the original expression:

\[ \log(2) + \log(3) - \log(6) \]

### Step 3: Use logarithmic properties

Using the property of logarithms that states \(\log(a) + \log(b) = \log(ab)\), we can combine the first two logarithms:

\[ \log(2) + \log(3) = \log(2 \cdot 3) = \log(6) \]

### Step 4: Simplify the expression

Now, substituting this back into our expression gives:

\[ \log(6) - \log(6) = 0 \]

### Final Result

Thus, the final answer is:

\[ \boxed{0} \]