Evaluate `lim_(xto0)(log(5+x)-log(5-x))/x`

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{\log(5+x) - \log(5-x)}{x}, \]

we can use properties of logarithms and limits. Here is the step-by-step solution:

### Step 1: Use the property of logarithms
We can combine the logarithmic expressions using the property that \(\log a - \log b = \log \left(\frac{a}{b}\right)\):

\[ \log(5+x) - \log(5-x) = \log\left(\frac{5+x}{5-x}\right). \]

Thus, we rewrite the limit as:

\[ \lim_{x \to 0} \frac{\log\left(\frac{5+x}{5-x}\right)}{x}. \]

### Step 2: Apply L'Hôpital's Rule
As \(x\) approaches 0, both the numerator and denominator approach 0, which gives us an indeterminate form of \(\frac{0}{0}\). Therefore, we can apply L'Hôpital's Rule:

\[ \lim_{x \to 0} \frac{\log\left(\frac{5+x}{5-x}\right)}{x} = \lim_{x \to 0} \frac{\frac{d}{dx}\left[\log\left(\frac{5+x}{5-x}\right)\right]}{\frac{d}{dx}[x]}. \]

### Step 3: Differentiate the numerator
Using the chain rule, we differentiate the numerator:

\[ \frac{d}{dx}\left[\log\left(\frac{5+x}{5-x}\right)\right] = \frac{1}{\frac{5+x}{5-x}} \cdot \frac{d}{dx}\left[\frac{5+x}{5-x}\right]. \]

Now, we differentiate \(\frac{5+x}{5-x}\):

\[ \frac{d}{dx}\left[\frac{5+x}{5-x}\right] = \frac{(5-x)(1) - (5+x)(-1)}{(5-x)^2} = \frac{5 - x + 5 + x}{(5-x)^2} = \frac{10}{(5-x)^2}. \]

Thus, we have:

\[ \frac{d}{dx}\left[\log\left(\frac{5+x}{5-x}\right)\right] = \frac{1}{\frac{5+x}{5-x}} \cdot \frac{10}{(5-x)^2} = \frac{10(5-x)}{(5+x)(5-x)^2}. \]

### Step 4: Substitute back into the limit
Now we substitute back into the limit:

\[ \lim_{x \to 0} \frac{\frac{10(5-x)}{(5+x)(5-x)^2}}{1} = \lim_{x \to 0} \frac{10(5-x)}{(5+x)(5-x)^2}. \]

### Step 5: Evaluate the limit
As \(x\) approaches 0, we can substitute \(x = 0\):

\[ \frac{10(5-0)}{(5+0)(5-0)^2} = \frac{10 \cdot 5}{5 \cdot 25} = \frac{50}{125} = \frac{2}{5}. \]

### Final Answer
Thus, the limit evaluates to:

\[ \lim_{x \to 0} \frac{\log(5+x) - \log(5-x)}{x} = \frac{2}{5}. \]