If `lim_(xrarr0) (log (3+x)-log (3-x))/(x)=k`, the value of k is

Options

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

Answer

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

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

**Hint:** Check if substituting the limit gives an indeterminate form like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).

### Step 2: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, we differentiate the numerator and the denominator separately:
\[ \text{Numerator: } \frac{d}{dx}[\log(3+x) - \log(3-x)] = \frac{1}{3+x} - \frac{1}{3-x} \]
\[ \text{Denominator: } \frac{d}{dx}[x] = 1 \]

### Step 3: Rewrite the limit using derivatives
Now we can rewrite the limit:
\[ \lim_{x \to 0} \frac{\frac{1}{3+x} - \frac{1}{3-x}}{1} \]

### Step 4: Simplify the expression
Now we 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, the limit becomes:
\[ \lim_{x \to 0} \frac{-2x}{(3+x)(3-x)} \]

### Step 5: Substitute \( x = 0 \)
Now we substitute \( x = 0 \) into the simplified limit:
\[ \lim_{x \to 0} \frac{-2x}{(3+x)(3-x)} = \frac{-2(0)}{(3+0)(3-0)} = \frac{0}{9} = 0 \]

### Step 6: Conclusion
Thus, we find that:
\[ k = 0 \]

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

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