Evaluate $\lim\limits_{n\rightarrow \infty}\left\{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\frac{3^3}{n^4}+\dots +\frac{n^3}{n^4}\right\}$ 
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*Evaluate $\displaystyle\lim_{n\rightarrow \infty}\left\{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\frac{3^3}{n^4}+\dots +\frac{n^3}{n^4}\right\}$.

*Examine whether $x^{1/x}$ possesses a maximum or minimum and determine the same. 
 A: It is happens that there is an attractive closed form for $1^3+2^3+\cdots+n^3$, which can readily be proved by induction:
$$1^3+2^3+3^3+\cdots +n^3=\left(\frac{n(n+1)}{2}\right)^2.$$
Divide by $n^4$. Fairly quickly we find that the desired limit is $\dfrac{1}{4}$. 
A: $${\frac{1^3}{n^4}+\frac{2^3}{n^4}+\frac{3^3}{n^4}+\dots +\frac{n^3}{n^4}}=\frac{\sum_{i=1}^{n}i^3}{n^4}=\frac{(n(n+1))^2}{4n^4}=\frac{1}{4}(1+\frac{1}{n})^2$$
$$\Rightarrow \displaystyle \lim_{n\to\infty}{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\frac{3^3}{n^4}+\dots +\frac{n^3}{n^4}}=\lim_{n\to\infty}\frac{1}{4}(1+\frac{1}{n})^2=1/4$$(Using the fact that $\lim_{n\to\infty}1/n=0)$
A: $\displaystyle\lim_{n\rightarrow \infty}\left\{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\frac{3^3}{n^4}+\dots +\frac{n^3}{n^4}\right\}$
= $\displaystyle\lim_{n\rightarrow \infty}\left\{\frac{1^3+2^3+3^3+\dots +n^3}{n^4}\right\}$  =
 $\displaystyle\lim_{n\rightarrow \infty}\left\{\left(\frac{n(n+1)}{2n^2}\right)^2\right\}$
=$\frac{1}{4}\displaystyle\lim_{n\rightarrow \infty}\left(1+\frac{1}{n^2} \right)^2$
=$\frac{1}{4}$
A: This begs to be viewed as a Riemann sum:
$$
\frac 1n\left(\frac{1^3}{n^3}+\cdots+\frac{n^3}{n^3}\right) \to \int_0^1 x^3\,dx.
$$
Your second question seems quite different from your first and should be posted separately.
A: Hints:
1) This is a Riemann sum.
2) Differentiate and solve for $x$ after equating the first derivative to zero, then try second derivative test to conclude whether the values of $x$ thus obtained is maxima or minima.
