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Miscellaneous 7 II · Q123

Q.lim⁡x→0[a3x−a2x−ax+1x⋅tan⁡x]\displaystyle\lim_{x\to 0}\left[\frac{a^{3x}-a^{2x}-a^{x}+1}{x\cdot\tan x}\right]

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a3x−a2x−ax+1=a2x(ax−1)−(ax−1)=(ax−1)(a2x−1)a^{3x}-a^{2x}-a^x+1=a^{2x}(a^x-1)-(a^x-1)=(a^x-1)(a^{2x}-1). Approximating, (ax−1)∼xlog⁡a(a^x-1)\sim x\log a and (a2x−1)∼2xlog⁡a(a^{2x}-1)\sim2x\log a, so the numerator ∼2x2(log⁡a)2\sim2x^2(\log a)^2. Denomina …

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