$\lim_{n\rightarrow\infty}\sqrt{n^6+n^2+1}-n\sqrt{n^4+2}=$

A cura di: Administrator

Limite in forma indeterminata.
Razionalizzando si ottiene

$lim_{nrightarrowinfty}sqrt{n^6+n^2+1}-nsqrt{n^4+2}=$

$lim_{nrightarrowinfty}(sqrt{n^6+n^2+1}-sqrt{n^6+2n^2})=$

$lim_{nrightarrowinfty}frac{sqrt{n^6+n^2+1}-sqrt{n^6+2n^2}}{sqrt{n^6+n^2+1}+sqrt{n^6+2n^2}}cdot sqrt{n^6+n^2+1}+sqrt{n^6+2n^2}=$

$ lim_{nrightarrowinfty}frac{n^6+n^2+1-(n^6+2n^2)}{sqrt{n^6+n^2+1}+sqrt{n^6+2n^2}} =$

$lim_{nrightarrowinfty}frac{-n^2+1}{sqrt{n^6cdot(1+frac{1}{n^4}+frac{1}{n^6})}-sqrt{n^6cdot(1+frac{2}{n^4})}} =$

$lim_{nrightarrowinfty}frac{n^2cdot(1-frac{1}{n^2})}{n^3cdot(sqrt{1+frac{1}{n^4}+frac{1}{n^6}}+sqrt{1+frac{2}{n^4}})} = 0$