Jawabanmu

2014-04-01T18:53:11+07:00
F(x) dimasukkan, kemudian f(1), x-nya diganti 1 semua. Sulit menulisnya disini.
 \lim_{x \to \21}  \frac{ x^{2}+4x-3-(1+4-3) }{x-1}  = \lim_{x \to \21}  \frac{ x^{2} +4x-3-2}{x-1} = \lim_{x \to \21}  \frac{ x^{2} +4x-5}{x-1}
 \lim_{x \to \21}  \frac{(x-1)(x+5)}{x-1}  = \lim_{x \to \21} x+5 = 1+5 = 6