Question

Lim când x—>0 din (sin x - x*cos x)/x^3

Answer 1

[tex]\text{Suntem in cazul de nedeterminare } \dfrac{0}{0}\text{ deci putem aplica l'Hopital.}\\ \displaystyle \limit\lim_{x\to 0}\dfrac{\sin x-x\cdot \cos x}{x^3}\stackrel{\frac{0}{0}}{=} \limit\lim_{x\to 0} \dfrac{\cos x-\cos x +x\cdot \sin x}{3x^2}=\limit\lim_{x\to 0} \dfrac{\sin x}{3x}= \dfrac{1}{3} [/tex]

Answer 2

$\lim_{x \to \ 0} \frac{sinx-xcosx}{x^{3}} =Aplicam~regula~lui~L'Hospital=$$\lim_{x \to \ 0} \frac{cosx-(x'cosx+x*(cosx)')}{(x^{3})'} = \lim_{x \to \ 0} \frac{cosx-cosx+xsinx}{3x^{2}} =$$\lim_{x \to \ 0} \frac{xsinx}{3x^{2}} = \lim_{x \to \ 0} \frac{sinx}{3x} =Aplica~regula~lui~L'Hospital=$$\lim_{x \to \ 0} \frac{(sinx)'}{3} = \lim_{x \to \ 0} \frac{cosx}{3} =cos0/3=1/3$