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Question Number 119852 by benjo_mathlover last updated on 27/Oct/20

$$\underset{n\to \mathrm{\infty }}{\lim }{n}^2\int \underset{0}{\stackrel{\frac{1}{n}}{}}{x}^{x+1}dx=?$$

Answered by Olaf last updated on 27/Oct/20

$${\int }_0^{\frac{1}{n}}{x}^{n+1}dx={\left[\frac{x^{n+2}}{n+2}\right]}_0^{\frac{1}{n}}=\frac{1}{(n+2)n^{n+2}}$$$$n^2{\int }_0^{\frac{1}{n}}{x}^{n+1}dx=\frac{1}{(n+2)n^n}\underset{\mathrm{\infty }}{\to }0$$$$$$

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