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A: underset(x to oo)"Lt" x sin""1/x=0 R: If underset(x to a)"lt" f(x)=0 and g(x) is bounded on a deleted neighbourhood ofa then underset(x to a)"Lt" f(x) g(x)=0 |
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Answer» Both A and R are TRUE and R is the CORRECT EXPLANATION of A |
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