第四小题求极限
2016-09-27
展开全部
(1)
=(2²+1)/(2*2²-1)
= 5/7
(2)
=2
(3)
=(x→-1)lim[(x+1)(x-1)]/[(x+4)(x+1)]
= (x→-1)lim(x-1)/(x+4)
= (-1-1)/(-1+4)
= -2/3
(4)
= 1×2
= 2
(5)
= (x→0)lim[(a²+2ax+x²-a²)/x]
= (x→0)lim[(2ax+x²)/x]
= (x→0)lim(2a+x)
= 2a
(6)
= (x→1)lim{(x+1-2)/[(x+1)(x-1)]}
= (x→1)lim{(x-1)/[(x+1)(x-1)]}
= (x→1)lim{1/[x+1]}
= 1/(1+1)
= 1/2
=(2²+1)/(2*2²-1)
= 5/7
(2)
=2
(3)
=(x→-1)lim[(x+1)(x-1)]/[(x+4)(x+1)]
= (x→-1)lim(x-1)/(x+4)
= (-1-1)/(-1+4)
= -2/3
(4)
= 1×2
= 2
(5)
= (x→0)lim[(a²+2ax+x²-a²)/x]
= (x→0)lim[(2ax+x²)/x]
= (x→0)lim(2a+x)
= 2a
(6)
= (x→1)lim{(x+1-2)/[(x+1)(x-1)]}
= (x→1)lim{(x-1)/[(x+1)(x-1)]}
= (x→1)lim{1/[x+1]}
= 1/(1+1)
= 1/2
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