`lim_(x->1)(x^2-x*lnx+lnx-1)/(x-1)`

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the limit \( \lim_{x \to 1} \frac{x^2 - x \ln x + \ln x - 1}{x - 1} \), we will follow these steps:

### Step 1: Substitute \( x = 1 \)
First, we substitute \( x = 1 \) into the expression to check if it results in an indeterminate form.

\[ \text{Numerator: } 1^2 - 1 \cdot \ln(1) + \ln(1) - 1 = 1 - 0 + 0 - 1 = 0 \]
\[ \text{Denominator: } 1 - 1 = 0 \]

Since both the numerator and denominator evaluate to 0, we have a \( \frac{0}{0} \) indeterminate form.

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule. This involves differentiating the numerator and the denominator separately.

**Numerator:**
\[ f(x) = x^2 - x \ln x + \ln x - 1 \]
Differentiating \( f(x) \):
1. The derivative of \( x^2 \) is \( 2x \).
2. For \( -x \ln x \), we use the product rule:
\[ \frac{d}{dx}(-x \ln x) = -\left(\ln x + 1\right) \]
3. The derivative of \( \ln x \) is \( \frac{1}{x} \).
4. The derivative of a constant (1) is 0.

Combining these, we get:
\[ f'(x) = 2x - (\ln x + 1) + \frac{1}{x} \]

Thus, the derivative of the numerator is:
\[ f'(x) = 2x - \ln x - 1 + \frac{1}{x} \]

**Denominator:**
\[ g(x) = x - 1 \]
Differentiating \( g(x) \):
\[ g'(x) = 1 \]

### Step 3: Apply the limit again
Now we apply the limit again:
\[ \lim_{x \to 1} \frac{f'(x)}{g'(x)} = \lim_{x \to 1} (2x - \ln x - 1 + \frac{1}{x}) \]

Substituting \( x = 1 \):
\[ = 2(1) - \ln(1) - 1 + \frac{1}{1} = 2 - 0 - 1 + 1 = 2 \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to 1} \frac{x^2 - x \ln x + \ln x - 1}{x - 1} = 2 \]