$\lim_{n\rightarrow\infty}\sqrt{n^2+n}\cdot[1-\cos(\frac{1}{n})]$

A cura di: Administrator

Il limite forma indeterminata$+inftycdot 0$.

Ricordando il limite notevole $lim_{xrightarrow 0}frac{1-cos x}{x^2}=frac{1}{2}$, si ha

$lim_{nrightarrowinfty}sqrt{n^2+n}cdot[1-cos(frac{1}{n})] =$

$lim_{nrightarrowinfty}frac{sqrt{n^2+n}}{n^2}cdotfrac{[1-cos(frac{1}{n})]}{frac{1}{n^2}} =$

Posto ora $frac{1}{n}=t Rightarrow trightarrow 0 text{per} nrightarrowinfty$ si ottiene

$lim_{nrightarrowinfty} frac{sqrt{n^2cdot(1+frac{1}{n})}}{n^2}cdotlim_{trightarrow 0}frac{1-cos t}{t^2}=$

$frac{1}{2}cdotlim_{nrightarrowinfty}frac{|n|cdotsqrt{1+frac{1}{n}}}{n^2}=$

$frac{1}{2}cdot lim_{nrightarrowinfty}frac{sqrt{1+frac{1}{n}}}{|n|} =$

$frac{1}{2}cdot 0 = 0$