Evaluate `lim_(xto1)(x^(3)-x^(2)logx+logx-1)/(x^(2)-1)`

Answer

To evaluate the limit

\[ \lim_{x \to 1} \frac{x^3 - x^2 \log x + \log x - 1}{x^2 - 1}, \]

we first substitute \(x = 1\):

\[ \frac{1^3 - 1^2 \log 1 + \log 1 - 1}{1^2 - 1} = \frac{1 - 0 + 0 - 1}{1 - 1} = \frac{0}{0}. \]

Since this is an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule or simplify the expression.

### Step 1: Factor the numerator

We start by rewriting the numerator:

\[ x^3 - x^2 \log x + \log x - 1 = (x^3 - 1) - (x^2 \log x - \log x). \]

Now, we can factor \(x^3 - 1\):

\[ x^3 - 1 = (x - 1)(x^2 + x + 1). \]

### Step 2: Rewrite the numerator

Next, we can factor out \(\log x\) from the second part:

\[ x^2 \log x - \log x = \log x (x^2 - 1) = \log x (x - 1)(x + 1). \]

Thus, the numerator becomes:

\[ (x - 1)(x^2 + x + 1) - \log x (x - 1)(x + 1). \]

### Step 3: Factor out \(x - 1\)

Now we can factor out \((x - 1)\):

\[ (x - 1) \left( (x^2 + x + 1) - \log x (x + 1) \right). \]

### Step 4: Rewrite the limit

Now, we can rewrite the limit:

\[ \lim_{x \to 1} \frac{(x - 1) \left( (x^2 + x + 1) - \log x (x + 1) \right)}{(x - 1)(x + 1)}. \]

We can cancel \((x - 1)\) from the numerator and denominator:

\[ \lim_{x \to 1} \frac{(x^2 + x + 1) - \log x (x + 1)}{x + 1}. \]

### Step 5: Substitute \(x = 1\)

Now we can substitute \(x = 1\):

\[ \frac{(1^2 + 1 + 1) - \log 1 (1 + 1)}{1 + 1} = \frac{3 - 0}{2} = \frac{3}{2}. \]

### Final Answer

Thus, the limit evaluates to:

\[ \boxed{\frac{3}{2}}. \]