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微积分求极限
lim(x趋向无穷)(x2+(cosx)2-1)/(x+sinx)2=
人气:330 ℃ 时间:2020-03-23 19:10:56
解答
lim(x->oo)(x^2+(cosx)^2-1)/(x+sinx)^2=lim(x->oo)(x^2-(sinx)^2)/(x+sinx)^2分子分母同除以xsinx得到:=lim(x->oo)(x/sinx-sinx/x)/(x/sinx+sinx/x+2)=lim(x->oo)(x/sinx+sinx/x+2-2-2sinx/x)/(x/sinx+sin...
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