`lim_(x to1)(1+logx-x)/(1-2x-x^2)` equals

Options

A. 1
B. 0
C. `-1`
D. `-1//2`

Answer

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

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

First, we will substitute \( x = 1 \) into the expression:

\[ 1 + \log(1) - 1 = 1 + 0 - 1 = 0 \]

And for the denominator:

\[ 1 - 2(1) - (1)^2 = 1 - 2 - 1 = -2 \]

So, we have:

\[ \frac{0}{-2} = 0 \]

### Conclusion

Thus, the limit is:

\[ \lim_{x \to 1} \frac{1 + \log x - x}{1 - 2x - x^2} = 0 \]