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7.6 Q.III · Q83

Q.lim⁡x→0[ax−bxsin⁡(4x)−sin⁡(2x)]\displaystyle\lim_{x\to 0}\left[\frac{a^x-b^x}{\sin(4x)-\sin(2x)}\right]

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✓ Free question

ax−bx=(ax−1)−(bx−1)∼x(log⁡a−log⁡b)=xlog⁡(a/b)a^x-b^x=(a^x-1)-(b^x-1)\sim x(\log a-\log b)=x\log(a/b). Using the sum-to-product identity, sin⁡4x−sin⁡2x=2cos⁡3xsin⁡x\sin4x-\sin2x=2\cos3x\sin x; as x→0x\to0, cos⁡3x→1\cos3x\to1 and sin⁡x∼x\sin x\sim x, so the denominator ∼2x\sim2x. The limit is xlog⁡(a/b)2x=12log⁡(a/b)\dfrac{x\log(a/b)}{2x}=\dfrac12\log(a/b).

✓Final answer

12log⁡(a/b)\dfrac{1}{2}\log(a/b)

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