If f(x) = lim _{n rightarrow infty} n^2 left( | vedclass

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(C) Let $m = \frac{1}{n}$. As $n \rightarrow \infty$,$m \rightarrow 0$.$f(x) = \lim _{m \rightarrow 0} \frac{1}{m^2} \left(x^m - x^{\frac{m}{m+1}}\rig

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$\frac{x^2}{2} \log x - \frac{x^2}{4} + C$

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(C) Let $m = \frac{1}{n}$. As $n \rightarrow \infty$,$m \rightarrow 0$.$f(x) = \lim _{m \rightarrow 0} \frac{1}{m^2} \left(x^m - x^{\frac{m}{m+1}}\right)$$f(x) = \lim _{m \rightarrow 0} \frac{x^{\frac{m}{m+1}}}{m^2} \left(x^{m - \frac{m}{m+1}} - 1\right) = \lim _{m \rightarrow 0} \frac{x^{\frac{m}{m+1}}}{m^2} \left(x^{\frac{m^2}{m+1}} - 1\right)$$f(x) = \lim _{m \rightarrow 0} x^{\frac{m}{m+1}} \cdot \left(\frac{x^{\frac{m^2}{m+1}} - 1}{\frac{m^2}{m+1}}\right) \cdot \frac{1}{m+1}$Using the standard limit $\lim _{t \rightarrow 0} \frac{a^t - 1}{t} = \log a$,we get:$f(x) = x^0 \cdot \log x \cdot \frac{1}{0+1} = \log x$Now,$\int x f(x) d x = \int x \log x d x$.Using integration by parts: $\int u v d x = u \int v d x - \int (u' \int v d x) d x$.Let $u = \log x$ and $v = x$.$\int x \log x d x = \log x \cdot \frac{x^2}{2} - \int \frac{1}{x} \cdot \frac{x^2}{2} d x$$= \frac{x^2}{2} \log x - \int \frac{x}{2} d x = \frac{x^2}{2} \log x - \frac{x^2}{4} + C$.

If $f(x) = \lim _{n \rightarrow \infty} n^2 \left(x^{\frac{1}{n}} - x^{\frac{1}{n+1}}\right), x > 0$,then $\int x f(x) d x =$

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