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Miscellaneous 7 II · Q128

Q.lim⁡x→∞[(2x+1)2(7x−3)3(5x+2)5]\displaystyle\lim_{x\to \infty}\left[\frac{(2x+1)^2(7x-3)^3}{(5x+2)^5}\right]

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Leading behaviour: (2x+1)2∼4x2(2x+1)^2\sim4x^2, (7x−3)3∼343x3(7x-3)^3\sim343x^3, (5x+2)5∼3125x5(5x+2)^5\sim3125x^5. Total power 2+3=52+3=5 matches the denominator's power 55, so a finite limit exists: $\dfr …

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