Jawabanmu

2014-05-08T21:58:10+07:00
1. lim  \frac{6}{ x^{2} -5x+6} -  \frac{x+3}{x-3}
   = lim  \frac{6}{(x-2)(x-3)}- \frac{x+3}{(x-3)}
   = lim  \frac{6-[(x+3)(x-2)]}{(x-2)(x-3)}
   = lim  \frac{6-( x^{2}+x-6) }{(x-2)(x-3)}
   = lim  \frac{6- x^{2} -x+6}{(x-2)(x-3)}
   =  \lim_{x \to \33}  \frac{- x^{2} -x+12}{(x-2)(x-3)}
   =  \lim_{x \to \33}  \frac{-(x+4)(x-3)}{(x-2)(x-3)}
   =  \lim_{x \to \33} \frac{-(x+4)}{(x-2)}
   =  \frac{-(3+4)}{(3-2)}
   =  \frac{-7}{1}=-7

2.  \lim_{x \to \--4}  \frac{3x+36}{ x^{2} +2x-8}- \frac{x}{x+4}
   =  \lim_{x \to \--4}  \frac{3x+36}{(x+4)(x-2)}- \frac{x}{x+4}
   =  \lim_{x \to \--4}  \frac{3x+36-x(x-2)}{(x-2)(x+4)}
   =  \lim_{x \to \--4}  \frac{3x+36- x^{2} +2x}{(x-2)(x+4)}
   =  \lim_{x \to \--4}  \frac{ -x^{2}+5x+36}{(x-2)(x+4)}
   =  \lim_{x \to \--4}  \frac{-(x+4)(x-9)}{(x+4)(x-2)}
   =  \lim_{x \to \--4}  \frac{-(x-9)}{x-2}
   =  \frac{-[(-4)-9]}{[(-4)-2]}
   =  \frac{-13}{-6} =  \frac{13}{6}

Maaf kalo salah. :v
mbak saya tanya kan ini = lim \frac{6}{(x-2)(x-3)}- \frac{x+3}{(x-3)}
kok bisa jadi ini gimana caranya ?
= lim \frac{6-[(x+3)(x-2)]}{(x-2)(x-3)}
yang soal kedua harusnya jawabannya -13/6