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Miscellaneous 7 I · Q110

Q.lim⁡x→0(x⋅log⁡(1+3x)(e3x−1)2)=\displaystyle\lim_{x\to 0}\left(\frac{x\cdot\log(1+3x)}{(e^{3x}-1)^2}\right)= (A) 1e9\dfrac{1}{e^9} (B) 1e3\dfrac{1}{e^3} (C) 19\dfrac{1}{9} (D) 13\dfrac{1}{3}

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log⁡(1+3x)∼3x\log(1+3x)\sim3x, so the numerator xlog⁡(1+3x)∼3x2x\log(1+3x)\sim3x^2. Also e3x−1∼3xe^{3x}-1\sim3x, so (e3x−1)2∼9x2(e^{3x}-1)^2\sim9x^2. The limit is $\dfrac{3x^ …

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