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7.6 Q.I · Q76

Q.lim⁡x→0(8sin⁡x−2tan⁡xe2x−1)\displaystyle\lim_{x\to 0}\left(\frac{8^{\sin x}-2^{\tan x}}{e^{2x}-1}\right)

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Since sin⁡x→0\sin x\to0 and tan⁡x→0\tan x\to0 as x→0x\to0 (both ∼x\sim x), 8sin⁡x−1∼(sin⁡x)log⁡8∼xlog⁡88^{\sin x}-1\sim(\sin x)\log8\sim x\log8 and 2tan⁡x−1∼(tan⁡x)log⁡2∼xlog⁡22^{\tan x}-1\sim(\tan x)\log2\sim x\log2. So the numerator ∼x(log⁡8−log⁡2)=xlog⁡4\sim x(\log8-\log2)=x\log4. The denominator $ …

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