Evaluate `(lim)_(x->0)((log)_e x)/(x-1)`

Explanation

To evaluate the limit \(\lim_{x \to 0} \frac{\log_e x}{x - 1}\), we can follow these steps:

### Step 1: Rewrite the expression
We start with the limit:
\[ \lim_{x \to 0} \frac{\log_e x}{x - 1} \]
To simplify this, we can manipulate the logarithm. We can express \(\log_e x\) in a different form by adding and subtracting 1 inside the logarithm:
\[ \log_e x = \log_e (x + 1 - 1) = \log_e((x + 1) - 1) \]

### Step 2: Use the property of logarithms
Using the property of logarithms, we can rewrite it as:
\[ \log_e((x + 1) - 1) = \log_e(x + 1) - \log_e(1) \]
Since \(\log_e(1) = 0\), we have:
\[ \log_e x = \log_e(x + 1) - 0 = \log_e(x + 1) \]
Thus, our limit becomes:
\[ \lim_{x \to 0} \frac{\log_e(x + 1) - 0}{x - 1} \]

### Step 3: Change the variable
To evaluate the limit as \(x\) approaches 0, we can change the variable. Let \(y = x - 1\), then as \(x \to 0\), \(y \to -1\). Therefore, we rewrite \(x\) in terms of \(y\):
\[ x = y + 1 \]
Substituting this into our limit gives:
\[ \lim_{y \to -1} \frac{\log_e(y + 1)}{y} \]

### Step 4: Evaluate the limit
Now we can evaluate the limit:
\[ \lim_{y \to -1} \frac{\log_e(y + 1)}{y} \]
As \(y\) approaches -1, \(y + 1\) approaches 0. We can apply L'Hôpital's Rule since we have an indeterminate form \(\frac{0}{0}\):
\[ \lim_{y \to -1} \frac{\log_e(y + 1)}{y} = \lim_{y \to -1} \frac{\frac{d}{dy}(\log_e(y + 1))}{\frac{d}{dy}(y)} \]
Calculating the derivatives:
\[ \frac{d}{dy}(\log_e(y + 1)) = \frac{1}{y + 1}, \quad \text{and} \quad \frac{d}{dy}(y) = 1 \]
Thus, we have:
\[ \lim_{y \to -1} \frac{\frac{1}{y + 1}}{1} = \lim_{y \to -1} \frac{1}{y + 1} = \frac{1}{0} \quad \text{(undefined)} \]
However, we can evaluate the limit directly as \(y\) approaches -1:
\[ \lim_{y \to -1} \frac{\log_e(y + 1)}{y} = \lim_{y \to -1} \frac{\log_e(0)}{-1} = 0 \]

### Final Answer
Thus, the limit evaluates to:
\[ \lim_{x \to 0} \frac{\log_e x}{x - 1} = 0 \]