lim x -> 3 1 - cos (x+3) / x2 + 6x +9 = tolong dijawab yah pake cara, thanks :)

[tex] \frac{1-(1- \frac{1}{2}sin^{2} {(x+3)}) }{(x+3)(x+3)} [/tex]
= [tex] \frac{ \frac{1}{2} sin^{2}(x+3)}{(x+3)(x+3)} [/tex]
= [tex] \frac{1}{2} \frac{sin x+3}{x+3} \frac{sin x+3 }{x+3} = \frac{1}{2} 1 .1 = \frac{1}{2} [/tex]
semoga membantu!;)