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用洛比达法则求极限limx→0[x(E^x+1)%2(E^x%1)]...

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貌似不需要罗比达法则 这个极限等于 e^0-1=1-1=0

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待续

洛比达法则,也就是上下分别求导,再求极限 1题=0,2题=2

lim(x->e) (lnx-1)/(x-e) (0/0) = lim(x->e) (1/x)/1 =1/e or expands lnx about e lnx = lne +(x-e)/e + (x-e)^2/e^2+... = 1+(x-e)/e + (x-e)^2/e^2+... (lnx-1)/(x-e) = [ 1+(x-e)/e + (x-e)^2/e^2+... - 1] /(x-e) = ((x-e)/e + (x-e)^2/e^2...

lim(x→∞)[(1+x)/x]^(ax) =lim(x→∞)([1+(1/x)]^(x) )^a =e^a; lim[1+(1/x)]^(x)=e是个重要的极限 ∫(-∞,a)te^t dt =∫(-∞,a)t d(e^t) =t·e^t|(-∞,a) - ∫(-∞,a)e^t dt =a·e^a - lim(t→-∞)t·e^t -e^t|(-∞,a) =a·e^a - lim(t→-∞)t·e^t -e^a 令u=-t,则...

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