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JEE Main 2003
Exam
JEE Main
Year
2003
Subject
Mathematics
Chapter
Limits

JEE Main 2003 · Mathematics · Limits

_ n arrow ∞ (1+2^4+3^4+ n^4)/(n^5)- _ n arrow ∞ (1+2^3+3^3+...

lim⁡n→∞1+24+34+…n4n5−lim⁡n→∞1+23+33+…n3n5\lim _{n \rightarrow \infty} \frac{1+2^4+3^4+\ldots n^4}{n^5}-\lim _{n \rightarrow \infty} \frac{1+2^3+3^3+\ldots n^3}{n^5}

Options

  1. A15\frac{1}{5}
  2. B130\frac{1}{30}
  3. CZero
  4. D14\frac{1}{4}

Correct answer

A. 15\frac{1}{5}

Step-by-step solution

Lim⁡n→∞{(1n)4+(2n)4+(3n)4+……….(nn)4}\operatorname{Lim}_{n \rightarrow \infty}\left\{\left(\frac{1}{n}\right)^4+\left(\frac{2}{n}\right)^4+\left(\frac{3}{n}\right)^4+\ldots \ldots \ldots .\left(\frac{n}{n}\right)^4\right\}-\operatorname{Lim}_{n \rightarrow \infty} \frac{1}{n}{1n4+23n4+…….⋅n3n4}\left\{\frac{1}{n^4}+\frac{2^3}{n^4}+\ldots \ldots . \cdot \frac{n^3}{n^4}\right\} ∫01(x)4dx−0=[x55]01=15\int_0^1(x)^4 d x-0=\left[\frac{x^5}{5}\right]_0^1=\frac{1}{5}
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