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Question: Answered & Verified by Expert
$\lim _{n \rightarrow \infty} \frac{1+2^4+3^4+\ldots n^4}{n^5}-\lim _{n \rightarrow \infty} \frac{1+2^3+3^3+\ldots n^3}{n^5}$
MathematicsLimitsJEE MainJEE Main 2003
Options:
  • A
    $\frac{1}{5}$
  • B
    $\frac{1}{30}$
  • C
    Zero
  • D
    $\frac{1}{4}$
Solution:
1166 Upvotes Verified Answer
The correct answer is:
$\frac{1}{5}$
$\operatorname{Lim}_{n \rightarrow \infty}\left\{\left(\frac{1}{n}\right)^4+\left(\frac{2}{n}\right)^4+\left(\frac{3}{n}\right)^4+\ldots \ldots \ldots .\left(\frac{n}{n}\right)^4\right\}-\operatorname{Lim}_{n \rightarrow \infty} \frac{1}{n}\left\{\frac{1}{n^4}+\frac{2^3}{n^4}+\ldots \ldots . \cdot \frac{n^3}{n^4}\right\}$
$\int_0^1(x)^4 d x-0=\left[\frac{x^5}{5}\right]_0^1=\frac{1}{5}$

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