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JEE MATHEMATICS MCQ: LIMITS

1 Easy
1 of 1
Application +4 | -1
lim⁡x→0x2sin⁡πx\lim\limits_{x \to 0} {x^2}\sin \frac{\pi }{x} is equal to
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Correct option: B
0
Option B (0) is correct because the sine function is bounded between -1 and 1, and the x2x^2 term approaches 0 as x approaches 0, making the entire expression approach 0.

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