lim _{n rightarrow infty}left[frac{1}{n}+frac | vedclass

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Question

(C) Given the limit: $L = \lim _{n \rightarrow \infty} \sum_{r=0}^{4n} \frac{n^2}{(n+r)^3}$ We can rewrite the general term as: $\frac{n^2}{(n+r)^3} =

Vedclass · Accepted Answer

$\frac{12}{25}$

Vedclass · Answer

(C) Given the limit: $L = \lim _{n \rightarrow \infty} \sum_{r=0}^{4n} \frac{n^2}{(n+r)^3}$ We can rewrite the general term as: $\frac{n^2}{(n+r)^3} = \frac{n^2}{n^3(1+\frac{r}{n})^3} = \frac{1}{n} \cdot \frac{1}{(1+\frac{r}{n})^3}$ Thus,the expression becomes: $L = \lim _{n \rightarrow \infty} \sum_{r=0}^{4n} \frac{1}{n} \cdot \frac{1}{(1+\frac{r}{n})^3}$ Using the definition of definite integral as the limit of a sum,where $\frac{r}{n} = x$ and $\frac{1}{n} = dx$,as $n \rightarrow \infty$,the range of $x$ is from $0$ to $4$: $L = \int_{0}^{4} \frac{1}{(1+x)^3} dx$ $L = \left[ \frac{(1+x)^{-2}}{-2} \right]_{0}^{4} = -\frac{1}{2} \left[ \frac{1}{(1+x)^2} \right]_{0}^{4}$ $L = -\frac{1}{2} \left( \frac{1}{25} - 1 \right) = -\frac{1}{2} \left( -\frac{24}{25} \right) = \frac{12}{25}$ Hence,option $(c)$ is correct.

$\lim _{n \rightarrow \infty}\left[\frac{1}{n}+\frac{n^2}{(n+1)^3}+\frac{n^2}{(n+2)^3}+\frac{n^2}{(n+3)^3}+\ldots+\frac{n^2}{(n+4n)^3}\right]=$

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