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limx→0sin−1x−tan−1xln(1+x3)=A\lim_{x\rightarrow 0} \dfrac{\sin^{-1} x -\tan^{-1} x}{\ln (1+x^{3} )} = Ax→0limln(1+x3)sin−1x−tan−1x=A
Find the value of 1A\dfrac{1}{A}A1.
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