Question

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

Answer 1

$sfdisplaystyle lim_{n to infty} n^2left(sqrt[sf n]{sf x} -sqrt[sf n+1]{sf x}right)=lim_{n to infty}n^2left(x^{frac{1}{n}}-x^{frac{1}{n+1}}right)=lim_{n to infty}n^2cdotunderbrace{sf x^{frac{1}{n+1}}}_{=1}cdotleft(x^{frac{1}{n(n+1)}}-1right)=\ \ \ =lim_{n to infty}underbrace{sf frac{x^big{sf frac{1}{n^2+n}}-1}{frac{1}{n^2+n}}}_{sf =ln x}cdotfrac{n^2}{n^2+n}=ln xlim_{n to infty}frac{n^2}{n^2+n}=ln xcdot 1=ln x$