If $f(x) = \lim_{n \rightarrow \infty}{n^{2}(x^{1/n} - x^{1/(n + 1)}}$, $x 0,\ then\ \int_{}^{}{x\ f(x)}dx$ is equal to

Question

If $f(x) = \lim_{n \rightarrow \infty}{n^{2}(x^{1/n} - x^{1/(n + 1)}}$, $x > 0,\ then\ \int_{}^{}{x\ f(x)}dx$ is equal to

Dr. Abubacker · Accepted Answer

(d)
$f(x) = {\lim_{n ightarrow \infty}{n^{2}\ x}}^{1/(n + 1)}\left\lbrack x^{\frac{1}{n} - \frac{1}{n + 1}} - 1\  ightbrack$
$= \lim_{n ightarrow \infty}\frac{x^{1/(n + 1)}\left( \frac{1}{x^{n(n + 1)}} - 1 ight)}{\frac{1}{n(n + 1)} 	imes \frac{n(n + 1)}{n^{2}}}\log{x = \log x}$
Hence, $\int_{}^{}{x\ f(x)dx = \int_{}^{}{x\log{x\ dx}}}$
$= \frac{x^{2}}{2}\log{x - \frac{1}{4}x^{2} + c}$

Q. If $f(x) = \lim_{n \rightarrow \infty}{n^{2}(x^{1/n} - x^{1/(n + 1)}}$, $x > 0,\ then\ \int_{}^{}{x\ f(x)}dx$ is equal to

Explanation

Reviewed by Dr. Abubacker

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