极限问题怎么解呢
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2019-02-12 · 中小学教师,杨建朝,蒲城县教研室蒲城县教育学会、教育领域创作...
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lim(x->-1) [1/(ax+1) - 3/(x^3+1) ] =b
lim(x->-1) [x^3-1 -3(ax+1) ]/[(ax+1).(x^3+1) ] =b
lim(x->-1) (x^3-3ax-4 )/[(ax+1).(x^3+1) ] =b (0/0)
(-1)^3 -3a(-1) -4 =0
a= 5/3
lim(x->-1) (x^3-3ax-4 )/[(ax+1).(x^3+1) ] =b
lim(x->-1) (x^3-5x-4 )/[((5/3)x+1).(x^3+1) ] =b
lim(x->-1) (x+1)(x^2-x-4) /[((5/3)x+1).(x+1)(x^2-x+1) ] =b
lim(x->-1) (x^2-x-4) /[((5/3)x+1)(x^2-x+1) ] =b
(1+1-4) /[ (-5/3 +1 ) ( 1 +1+1) ] =b
b=1
(a,b) = (5/3 ,1 )
lim(x->-1) [x^3-1 -3(ax+1) ]/[(ax+1).(x^3+1) ] =b
lim(x->-1) (x^3-3ax-4 )/[(ax+1).(x^3+1) ] =b (0/0)
(-1)^3 -3a(-1) -4 =0
a= 5/3
lim(x->-1) (x^3-3ax-4 )/[(ax+1).(x^3+1) ] =b
lim(x->-1) (x^3-5x-4 )/[((5/3)x+1).(x^3+1) ] =b
lim(x->-1) (x+1)(x^2-x-4) /[((5/3)x+1).(x+1)(x^2-x+1) ] =b
lim(x->-1) (x^2-x-4) /[((5/3)x+1)(x^2-x+1) ] =b
(1+1-4) /[ (-5/3 +1 ) ( 1 +1+1) ] =b
b=1
(a,b) = (5/3 ,1 )
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