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用泰勒展开式.
√ln(x+1)=√(x-x^2/2+o(x^2))
所以√x-√ln(x+1)=√x-√(x-x^2/2+o(x^2))
=(x-x+x^2/2+o(x^2))/(√x+√(x-x^2/2+o(x^2))
=(x^2/2+o(x^2))/(√x+√(x-x^2/2+o(x^2))
所以lim(x→0)(√x-√ln(x+1))/(cx^k)=lim(x→0)(x^2/2+o(x^2))/(cx^k(√x+√(x-x^2/2+o(x^2)))
=lim(x→0)(1/2+o(1))/(cx^(k-3/2)(1+√(1-x/2+o(x)))=1
所以k-3/2=0,k=3/2
所以(1/2)/(c(1+1))=1,c=1/4
√ln(x+1)=√(x-x^2/2+o(x^2))
所以√x-√ln(x+1)=√x-√(x-x^2/2+o(x^2))
=(x-x+x^2/2+o(x^2))/(√x+√(x-x^2/2+o(x^2))
=(x^2/2+o(x^2))/(√x+√(x-x^2/2+o(x^2))
所以lim(x→0)(√x-√ln(x+1))/(cx^k)=lim(x→0)(x^2/2+o(x^2))/(cx^k(√x+√(x-x^2/2+o(x^2)))
=lim(x→0)(1/2+o(1))/(cx^(k-3/2)(1+√(1-x/2+o(x)))=1
所以k-3/2=0,k=3/2
所以(1/2)/(c(1+1))=1,c=1/4
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