Question

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$\lim_{n \to \infty} n^2(\sqrt[n]{x} -\sqrt[n+1]{x})$
$x\ \textgreater \ 0$

Answer 1

$\sf\displaystyle \lim_{n \to \infty} n^2\left(\sqrt[\sf n]{\sf x} -\sqrt[\sf n+1]{\sf x}\right)=\lim_{n \to \infty}n^2\left(x^{\frac{1}{n}}-x^{\frac{1}{n+1}}\right)=\lim_{n \to \infty}n^2\cdot\underbrace{\sf x^{\frac{1}{n+1}}}_{=1}\cdot\left(x^{\frac{1}{n(n+1)}}-1\right)=\\ \\ \\ =\lim_{n \to \infty}\underbrace{\sf \frac{x^\big{\sf \frac{1}{n^2+n}}-1}{\frac{1}{n^2+n}}}_{\sf =\ln x}\cdot\frac{n^2}{n^2+n}=\ln x\lim_{n \to \infty}\frac{n^2}{n^2+n}=\ln x\cdot 1=\ln x$