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Limits — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
lim⁡x→∞∑r=110(x+r)2010(x1006+1)(2x1004+1) \displaystyle\lim_{x\to\infty}\dfrac{\sum_{r=1}^{10}(x+r)^{2010}}{(x^{1006}+1)(2x^{1004}+1)}\, is equal to
A55
B20102010
C12\dfrac{1}{2}correct
D00
Solution
Step 1: Highest power in numerator. Each (x+r)2010(x+r)^{2010} has leading term x2010x^{2010}. Summing 10 such terms gives 10 x2010+10\,x^{2010}+ lower powers. Step 2: Highest power in denominator: (x1006+1)(2x1004+1)=2x2010+(x^{1006}+1)(2x^{1004}+1)=2x^{2010}+ lower powers. Step 3: Both have degree 20102010. The limit equals the ratio of leading coefficients:
lim⁡x→∞10 x2010+…2 x2010+…=102=5\lim_{x\to\infty}\dfrac{10\,x^{2010}+\ldots}{2\,x^{2010}+\ldots}=\dfrac{10}{2}=5
 Correct answer: (A) 55
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