lim(x→-2)(ax+b)/(x+2)=4,a、b为常数,求a、b.
问题描述:
lim(x→-2)(ax+b)/(x+2)=4,a、b为常数,求a、b.
答
……
答
分子在x=-2时为零,故-2a+b=0,且左边的极限为a/1=4,故b=8,a=4
答
lim(x→-2)(ax+b)/(x+2)=lim(x→-2)[a(x+2)+b-2a]/(x+2)=a+lim(x→-2)(b-2a)/(x+2)由于lim(x→-2)1/(x+2)=∞所以lim(x→-2)(b-2a)/(x+2)为常数的条件是b-2a=0,即b=2a那么lim(x→-2)(ax+b)/(x+2)=a=4,故b=8...