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7.6 Q.I · Q75

Q.lim⁡x→0(6x+5x+4x−3x+1sin⁡x)\displaystyle\lim_{x\to 0}\left(\frac{6^x+5^x+4^x-3^{x+1}}{\sin x}\right)

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6x+5x+4x−3x+1=6x+5x+4x−3⋅3x=(6x−1)+(5x−1)+(4x−1)−3(3x−1)6^x+5^x+4^x-3^{x+1}=6^x+5^x+4^x-3\cdot3^x=(6^x-1)+(5^x-1)+(4^x-1)-3(3^x-1) (the constants 1+1+1−3=01+1+1-3=0 check out). Dividing by xx: $\log6+\log5+\log4-3\log3=\log6+\log5+\log4-\log27=\log\dfrac{6\times5\times4}{27}=\log\dfrac{120}{27}=\ …

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