`Lt_(x to 0) (log(3+x) - log (3-x))/x =k` then the value of :

Options

A. 0
B. `-1/3`
C. `2/3`
D. `-2/3`

Answer

To solve the limit \( \lim_{x \to 0} \frac{\log(3+x) - \log(3-x)}{x} = k \), we can follow these steps:

### Step 1: Identify the form of the limit
When we substitute \( x = 0 \) directly into the limit, we get:
\[ \frac{\log(3+0) - \log(3-0)}{0} = \frac{\log(3) - \log(3)}{0} = \frac{0}{0} \]
This is an indeterminate form, so we can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
L'Hôpital's Rule states that if we have an indeterminate form \( \frac{0}{0} \), we can take the derivative of the numerator and the derivative of the denominator:
\[ \text{Numerator: } \frac{d}{dx}[\log(3+x) - \log(3-x)] = \frac{1}{3+x} - \frac{1}{3-x} \]
\[ \text{Denominator: } \frac{d}{dx}[x] = 1 \]
Now we can rewrite the limit:
\[ \lim_{x \to 0} \frac{\frac{1}{3+x} - \frac{1}{3-x}}{1} \]

### Step 3: Simplify the expression
Now we need to simplify the expression in the limit:
\[ \frac{1}{3+x} - \frac{1}{3-x} = \frac{(3-x) - (3+x)}{(3+x)(3-x)} = \frac{-2x}{(3+x)(3-x)} \]
Thus, we have:
\[ \lim_{x \to 0} \frac{-2x}{(3+x)(3-x)} \]

### Step 4: Substitute \( x = 0 \)
Now we substitute \( x = 0 \) into the simplified expression:
\[ \lim_{x \to 0} \frac{-2x}{(3+x)(3-x)} = \frac{-2(0)}{(3+0)(3-0)} = \frac{0}{9} = 0 \]

### Step 5: Conclusion
Thus, we find that:
\[ k = 0 \]

### Final Answer:
The value of \( k \) is \( 0 \).

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