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7.6 Q.III · Q86

Q.lim⁡x→0[(25)x−2(5)x+1x⋅sin⁡x]\displaystyle\lim_{x\to 0}\left[\frac{(25)^x-2(5)^x+1}{x\cdot \sin x}\right]

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(5x−1)2=25x−2⋅5x+1(5^x-1)^2=25^x-2\cdot5^x+1, matching the numerator. So the numerator ∼(xlog⁡5)2=x2(log⁡5)2\sim(x\log5)^2=x^2(\log5)^2. The denominator $x\sin x\sim …

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