A cura di: Administrator

Il limite si presenta in forma indeterminata ($inftycdot 0$).

Si ha

$lim_{nrightarrowinfty} n^2 sin(frac{1}{n}) sqrt{1-cos(frac{2}{n})}=$

$lim_{nrightarrowinfty} frac{sin(frac{1}{n})}{frac{1}{n}} frac{sqrt{1-cosfrac{2}{n}}}{sqrt{(frac{1}{n})^2}}=$

$lim_{nrightarrowinfty} frac{sin(frac{1}{n})}{frac{1}{n}} sqrt{frac{1-cos(frac{2}{n})}{frac{1}{4} (frac{2}{n})^2}}$

Posto allora $frac{1}{n} = t$ e $frac{2}{n}=k$, risulta

$lim_{nrightarrowinfty} frac{sin(frac{1}{n})}{frac{1}{n}} 2 sqrt{frac{1-cos(frac{2}{n})}{(frac{2}{n})^2}} =$

$2cdotlim_{trightarrow 0}frac{sin t}{t}cdotlim_{krightarrow 0} sqrt{frac{1-cos k}{k^2}}=2cdot 1cdotfrac{1}{sqrt{2}}=sqrt{2}$