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Miscellaneous 7 I · Q113

Q.lim⁡x→∞[(2x+3)7(x−5)3(2x−5)10]=\displaystyle\lim_{x\to \infty}\left[\frac{(2x+3)^7(x-5)^3}{(2x-5)^{10}}\right]= (A) 38\dfrac{3}{8} (B) 18\dfrac{1}{8} (C) 16\dfrac{1}{6} (D) 14\dfrac{1}{4}

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Leading behaviour: (2x+3)7∼(2x)7=128x7(2x+3)^7\sim(2x)^7=128x^7, (x−5)3∼x3(x-5)^3\sim x^3, (2x−5)10∼(2x)10=1024x10(2x-5)^{10}\sim(2x)^{10}=1024x^{10}. Total numerator power 7+3=107+3=10, matching the denominator, so a finite limit exists: $\dfrac{ …

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