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7.6 Q.III · Q87

Q.lim⁡x→0[(49)x−2(35)x+(25)xsin⁡x⋅log⁡(1+2x)]\displaystyle\lim_{x\to 0}\left[\frac{(49)^x-2(35)^x+(25)^x}{\sin x\cdot\log(1+2x)}\right]

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(7x−5x)2=49x−2⋅35x+25x(7^x-5^x)^2=49^x-2\cdot35^x+25^x, matching the numerator (since 35x=7x⋅5x35^x=7^x\cdot5^x). And 7x−5x∼x(log⁡7−log⁡5)=xlog⁡(7/5)7^x-5^x\sim x(\log7-\log5)=x\log(7/5), so the numerator ∼x2[log⁡(7/5)]2\sim x^2[\log(7/5)]^2. Denominator: $\sin x\cdot\log(1+2x)\sim x\tim …

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