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$\lim _{x \rightarrow \infty}\left[\frac{1^3+2^3+3^3+\ldots \ldots .+n^3}{n^4}\right]=$
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The correct answer is:
$\frac{1}{4}$
\(\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{1^3+2^3+3^3+\ldots+n^3}{n^4} \\
& =\lim _{n \rightarrow \infty} \frac{\left[\frac{1}{2} n(n+1)\right]^2}{n^4} \\
& {\left[1^3+2^3+3^3+\ldots+n^3=\left(\frac{1}{2} n(n+1)\right)^2\right]} \\
& =\lim _{n \rightarrow \infty} \frac{\frac{1}{4} n^2(n+1)^2}{n^4} \\
& =\lim _{n \rightarrow \infty} \frac{\frac{1}{4} n^2\left(n^2+1+2 n\right)}{n^4} \\
& =\lim _{4 n \rightarrow \infty} \frac{n^4\left(1+\frac{1}{n^2}+\frac{2}{n}\right)}{n^4} \quad \quad\left[\frac{\infty}{\infty} \text { form }\right] \\
& =\frac{1}{4} \lim _{n \rightarrow \infty} \frac{\left(1+\frac{1}{n^2}+\frac{2}{n}\right)}{1}
\end{aligned}\)
When \(\mathrm{n} \rightarrow \infty\), then \(\frac{1}{\mathrm{n}} \rightarrow 0=\frac{1}{4}\)
& \lim _{n \rightarrow \infty} \frac{1^3+2^3+3^3+\ldots+n^3}{n^4} \\
& =\lim _{n \rightarrow \infty} \frac{\left[\frac{1}{2} n(n+1)\right]^2}{n^4} \\
& {\left[1^3+2^3+3^3+\ldots+n^3=\left(\frac{1}{2} n(n+1)\right)^2\right]} \\
& =\lim _{n \rightarrow \infty} \frac{\frac{1}{4} n^2(n+1)^2}{n^4} \\
& =\lim _{n \rightarrow \infty} \frac{\frac{1}{4} n^2\left(n^2+1+2 n\right)}{n^4} \\
& =\lim _{4 n \rightarrow \infty} \frac{n^4\left(1+\frac{1}{n^2}+\frac{2}{n}\right)}{n^4} \quad \quad\left[\frac{\infty}{\infty} \text { form }\right] \\
& =\frac{1}{4} \lim _{n \rightarrow \infty} \frac{\left(1+\frac{1}{n^2}+\frac{2}{n}\right)}{1}
\end{aligned}\)
When \(\mathrm{n} \rightarrow \infty\), then \(\frac{1}{\mathrm{n}} \rightarrow 0=\frac{1}{4}\)
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