A cura di: Administrator

Il limite si presenta in forma indeterminata. $lim_{xrightarrow pi/2}frac{sin 4xcdotsin 3x}{xcdotsin 2x}$

Posto $x-frac{pi}{2} = t, allora trightarrow 0$ per $xrightarrowfrac{pi}{2}$ e risulta

$lim_{trightarrow 0}frac{sin[4(t+frac{pi}{2})]cdotsin[3(t+frac{pi}{2})]}{(t+frac{pi}{2})cdotsin[2(t+frac{pi}{2})]} = lim_{trightarrow 0}frac{sin(4t+2pi)cdotsin(3t+frac{3}{2}pi)}{(t+frac{pi}{2})cdotsin(2t+pi)} =$

 

$lim_{trightarrow 0}frac{sin 4tcdot(-cos 3t)}{(t+frac{pi}{2})cdot(-sin 2t)} = lim_{trightarrow 0}[frac{sin 4t}{4t}cdotfrac{2t}{-sin 2t}cdotfrac{2(-cos 3t)}{(t+frac{pi}{2})}] =$$= 1cdot(-1)cdotfrac{-2}{frac{pi}{2}} =frac{4}{pi}$