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Miscellaneous 7 I · Q106

Q.lim⁡x→0(15x−3x−5x+1sin⁡2x)=\displaystyle\lim_{x\to 0}\left(\frac{15^x-3^x-5^x+1}{\sin^2x}\right)= (A) log⁡15\log 15 (B) log⁡3+log⁡5\log 3+\log 5 (C) log⁡3.log⁡5\log 3.\log 5 (D) 3log⁡53\log 5

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(3x−1)(5x−1)=15x−3x−5x+1(3^x-1)(5^x-1)=15^x-3^x-5^x+1, matching the numerator. So the numerator ∼x2log⁡3log⁡5\sim x^2\log3\log5, and sin⁡2x∼x2\sin^2x\sim x^2. The limit is $\ …

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