The value of `lim_(x to infty) ((x^(2) sin (1//x) -x)/(1-|x|))` is:

Options

A. 0
B. 1
C. `-1`
D. none of these

Answer

To solve the limit \( \lim_{x \to \infty} \frac{x^2 \sin\left(\frac{1}{x}\right) - x}{1 - |x|} \), we will follow these steps:

### Step 1: Analyze the expression
As \( x \) approaches infinity, we need to simplify the expression. Notice that \( |x| = x \) when \( x \) is positive. Therefore, the expression can be rewritten as:
\[ \lim_{x \to \infty} \frac{x^2 \sin\left(\frac{1}{x}\right) - x}{1 - x} \]

### Step 2: Simplify the numerator
We can factor out \( x \) from the numerator:
\[ = \lim_{x \to \infty} \frac{x \left(x \sin\left(\frac{1}{x}\right) - 1\right)}{1 - x} \]

### Step 3: Evaluate \( \sin\left(\frac{1}{x}\right) \)
As \( x \to \infty \), \( \frac{1}{x} \to 0 \). We know that \( \sin(t) \approx t \) when \( t \) is close to 0. Therefore:
\[ \sin\left(\frac{1}{x}\right) \approx \frac{1}{x} \]
Thus, we can substitute this approximation into our limit:
\[ x \sin\left(\frac{1}{x}\right) \approx x \cdot \frac{1}{x} = 1 \]

### Step 4: Substitute back into the limit
Now substituting back into our limit gives:
\[ = \lim_{x \to \infty} \frac{x(1 - 1)}{1 - x} = \lim_{x \to \infty} \frac{x \cdot 0}{1 - x} = \lim_{x \to \infty} \frac{0}{1 - x} = 0 \]

### Conclusion
Thus, the value of the limit is:
\[ \boxed{0} \]