急急急！！求数学能手，帮忙解题！
2024-03-14 16:51
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2024-03-14 21:15
lim(4x�0�5+2x+1)/(3x-1)=6 cosx=cos�0�5（x/2）-sin�0�5（x/2）lim[cos(x/2)+sin(x/2)]=√2 f（x）在x=-3无意义 连续，左右极限相等
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一、1. lim(8x^3-1)/(6x^2-5x+1)=lim24x^2/(12x-5)=24*(1/2)^2/(12*(1/2)-5)=62. limcosx/[cos(x/2)-sin(x/2)]=limcosx[cos(x/2)+sin(x/2)]/[(cos(x/2))^2-(sin(x/2))^2]=limcosx[cos(x/2)+sin(x/2)]/cosx=lim[cos(x/2)+sin(x/2)]=√2二、1. f(x)=(x^2-4)/[(x+2)(x+3)]所以在x=-3无定义lim(x→-3)(x+2)(x-2)/[(x+2)(x+3)]=lim(x-2)/(x+3)=∞所以x=-3是f(x)的第二类间断点2. lim(x从1的左侧趋于1)x^2=1=f(1)lim(x从1的右侧趋于1）（2-x)=1=f(1)所以f(x)在x=1连续
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