Question

calcule o limite : lim de X tendendo a 0 tg5x/sen6x

Answer 1

Por L'Hôpital:
$\displaystyle \lim_{x\to0}\frac{\tan (5x)}{\sin(6x)}=\lim_{x\to0}\frac{(\tan (5x))'}{(\sin(6x))'}\\\\ \frac{d}{dx}\tan(5x)=5.\sec^2(5x)\\\\\frac{d}{dx}\sin(6x)=6.\cos(6x)\\\\ \lim_{x\to0}\frac{5.\sec^2(5x)}{6.\cos(6x)}=\lim_{x\to0}\frac{5}{6\cos^2(5x).\cos(6x)}=\frac{5}{6\cos^2(5.0)\cos(6.0)}}\\\\ \implies \frac{5}{6.cos^2(0)\cos(0)}=\frac{5}{6.1.1}=\frac{5}{6}=\underline{0,8\bar{3}}$