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7.7 III · Q95

Q.lim⁡x→∞[(3x−4)3(4x+3)4(3x+2)7]\displaystyle\lim_{x\to \infty}\left[\frac{(3x-4)^3(4x+3)^4}{(3x+2)^7}\right]

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Leading behaviour: (3x−4)3∼(3x)3=27x3(3x-4)^3\sim(3x)^3=27x^3, (4x+3)4∼(4x)4=256x4(4x+3)^4\sim(4x)^4=256x^4, (3x+2)7∼(3x)7=2187x7(3x+2)^7\sim(3x)^7=2187x^7. So the ratio →27×256 x72187 x7=69122187\to\dfrac{27\times256\,x^7}{2187\,x^7}=\dfrac{6912}{2187}. Simplifying by the common factor 2727: 25681\dfrac{256}{81}.

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25681\dfrac{256}{81}

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