The value of `lim_(xrarroo) {(x^2sin((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 limit as \( x \to \infty \)
As \( x \) approaches infinity, \( |x| = x \). Therefore, we can rewrite the limit as:
\[ \lim_{x \to \infty} \frac{x^2 \sin\left(\frac{1}{x}\right) - x}{1 - x} \]

### Step 2: Simplify the expression
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: Rewrite the limit
Now, we can rewrite the limit as:
\[ = \lim_{x \to \infty} \frac{x \sin\left(\frac{1}{x}\right) - 1}{\frac{1 - x}{x}} = \lim_{x \to \infty} \frac{x \sin\left(\frac{1}{x}\right) - 1}{-\left(1 - \frac{1}{x}\right)} \]

### Step 4: Evaluate \( \sin\left(\frac{1}{x}\right) \)
As \( x \to \infty \), \( \frac{1}{x} \to 0 \). We can use the small angle approximation for sine:
\[ \sin\left(\frac{1}{x}\right) \approx \frac{1}{x} \]
Thus,
\[ x \sin\left(\frac{1}{x}\right) \approx x \cdot \frac{1}{x} = 1 \]

### Step 5: Substitute back into the limit
Now substituting this back, we get:
\[ = \lim_{x \to \infty} \frac{1 - 1}{-\left(1 - \frac{1}{x}\right)} = \lim_{x \to \infty} \frac{0}{-1} = 0 \]

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

---