Respuesta :

konrad509
[tex]\displaystyle \lim_{\theta\to 0}\dfrac{1-\cos \theta}{2\sin ^2\theta}=\\\\ \lim_{\theta\to 0}\dfrac{(1-\cos \theta)(1+\cos \theta)}{2\sin ^2\theta(1+\cos \theta)}=\\\\ \lim_{\theta\to 0}\dfrac{1-\cos^2 \theta}{2\sin ^2\theta+2\sin^2 \theta\cos \theta}=\\\\ \lim_{\theta\to 0}\dfrac{\sin^2 \theta}{2\sin ^2\theta+2\sin^2 \theta\cos \theta}=\\\\ \lim_{\theta\to 0}\dfrac{1}{2+2\cos \theta}=\\\\ \dfrac{1}{2+2\cdot1}=\\\\ \dfrac{1}{4} [/tex]

Otras preguntas