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$\underset{x \to \infty}{L i m} (\frac{x^{2} \sin (\frac{1}{x}) - x}{1 - | x |}) =$

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Correct option is A

Limx→∞x2sin(1x)−x1−|x|=Limx→∞x sin(1x)−11x−1 as x→∞⇒1x→0 Limx→∞1-10-1=0

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Limx→∞(x2sin(1x)−x1−|x|)=

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$\underset{x \to \infty}{L i m} (\frac{x^{2} \sin (\frac{1}{x}) - x}{1 - | x |}) =$