Question

Ajutatima va rog frumos cit se poate de repede

Answer 1

   
[tex]\displaystyle\\ 2)\\ \lim_{x \to 0} \frac{ \dfrac{1}{2}\sin 5x-3\sin \dfrac{x}{2}}{\sin 2x-4\sin \dfrac{3x}{2}+\sin 6x} \\\\ \texttt{Suntem in cazul: } \frac{0}{0} \\\\ \texttt{\bf Vom rezolva limita folosind formula lui L'Hopital.}\\\\ \lim_{x \to 0} \frac{ \dfrac{1}{2}\sin 5x-3\sin \dfrac{x}{2}}{\sin 2x-4\sin \dfrac{3x}{2}+\sin 6x}= \lim_{x \to 0} \frac{ \left(\dfrac{1}{2}\sin 5x-3\sin \dfrac{x}{2}\right)'}{\left(\sin 2x-4\sin \dfrac{3x}{2}+\sin 6x\right)'}= [/tex]


[tex]\displaystyle\\ =\lim_{x \to 0} \frac{\dfrac{5}{2}\cos 5x-\dfrac{3}{2}\cos \dfrac{x}{2}}{2\cos 2x-\dfrac{4\cdot 3}{2}\cos \dfrac{3x}{2}+6\cos 6x}=\\\\\\ =\frac{\dfrac{5}{2}\cdot 1-\dfrac{3}{2}\cdot 1}{2\cdot 1-\dfrac{4\cdot 3}{2}\cdot 1+6\cdot 1}=\frac{\dfrac{5}{2}-\dfrac{3}{2}}{2-6+6}=\frac{\dfrac{2}{2}}{2}= \boxed{\bf \frac{1}{2}} [/tex]