The value of `lim_(x to 0) (log (5 + x) - log (5 - x))/(x)` equals

Options

A. `(1)/(5)`
B. `(2)/(5)`
C. 0
D. Does not exist

Answer

To solve the limit \( \lim_{x \to 0} \frac{\log(5 + x) - \log(5 - x)}{x} \), we can follow these steps:

### Step 1: Write the limit expression
We start with the limit expression:
\[ \lim_{x \to 0} \frac{\log(5 + x) - \log(5 - x)}{x} \]

### Step 2: Check the form of the limit
Substituting \( x = 0 \) directly into the expression gives:
\[ \frac{\log(5 + 0) - \log(5 - 0)}{0} = \frac{\log(5) - \log(5)}{0} = \frac{0}{0} \]
This is an indeterminate form, so we can apply L'Hôpital's Rule.

### Step 3: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, we differentiate the numerator and the denominator separately:
- The derivative of the numerator \( \log(5 + x) - \log(5 - x) \) is:
\[ \frac{d}{dx}[\log(5 + x)] - \frac{d}{dx}[\log(5 - x)] = \frac{1}{5 + x} - \left(-\frac{1}{5 - x}\right) = \frac{1}{5 + x} + \frac{1}{5 - x} \]
- The derivative of the denominator \( x \) is \( 1 \).

Thus, we can rewrite the limit as:
\[ \lim_{x \to 0} \left( \frac{1}{5 + x} + \frac{1}{5 - x} \right) \]

### Step 4: Evaluate the limit
Now we can directly substitute \( x = 0 \):
\[ \lim_{x \to 0} \left( \frac{1}{5 + 0} + \frac{1}{5 - 0} \right) = \frac{1}{5} + \frac{1}{5} = \frac{2}{5} \]

### Final Answer
Thus, the value of the limit is:
\[ \frac{2}{5} \]

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