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Limits — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSLimits5 Marks
Question
Evaluate the following limit:
$\lim\limits_{\text{n}\rightarrow\infty}\frac{1^3+2^3+\ \dots+\text{n}^3}{(\text{n}-1)^4}$
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Answer

$\lim\limits_{\text{n}\rightarrow\infty}\frac{1^3+2^3+3^3\ \dots+\text{n}^3}{(\text{n}-1)^4}$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{\Big[\frac12(\text{n})(\text{n}+1)\Big]^2}{(\text{n}-1)^4}$ $\bigg[1^3+2^3+3^3+\ \cdots+\text{n}^3=\Big(\frac12\text{n}(\text{n}+1)\Big)^2\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\Bigg(\frac{\frac14\text{n}^2\big(\text{n}^2+1+2\text{n}\big)}{(\text{n}-1)^4}\Bigg)$
$=\frac14\lim\limits_{\text{n}\rightarrow\infty}\bigg(\frac{\text{n}^4+\text{n}^2+2\text{n}^3}{(\text{n}-1)^2(\text{n}-1)^2}\bigg)$ $\Big[\frac{\infty}{\infty}\text{ from}\Big]$
$=\frac{1}{4}\lim\limits_{\text{n}\rightarrow{\infty}}\Big(\frac{\text{n}^4+\text{n}^2+2\text{n}^3}{\text{n}^4+\text{n}^2-2\text{n}^3+\text{n}^2+1-2\text{n}-2\text{n}^3-2\text{n}+4\text{n}^2}\Big)$
$=\frac{1}{4}\lim\limits_{\text{n}\rightarrow{\infty}}\frac{\Big(1+\frac{1}{​​\text{n}^2}+\frac{2}{​​\text{n}}\Big)}{\Big(1+\frac{1}{\text{n}^2}+\frac{2}{\text{n}}+\frac{1}{\text{n}^2}+\frac{1}{\text{n}^4}-\frac{2}{\text{n}^3}-\frac{2}{\text{n}}-\frac{2}{\text{n}^3}+\frac{4}{\text{n}^2}\Big)}$
$=\frac{1}{4}\Big(\frac14\Big)$
$=\frac14$

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