√x-√in(1+x)~cx^k求c和k

Question

√x-√in(1+x)~cx^k求c和k

Answer 1

√x-√ln(1+x)~Cx^k
lim(x→0⁺)[√x-√ln(1+x)]/x^k=C
lim(x→0⁺)[√x-√ln(1+x)]/x^k
=lim(x→0⁺){1-√[ln(1+x)/x]}/x^(k-1/2)
=lim(x→0⁺)(-1/2)*1/√[ln(1+x)/x]*[ln(1+x)/x]'/[(k-1/2)*x^(k-3/2)]（洛必达法则）
=(-1/2)*1/(k-1/2)*lim(x→0⁺)[ln(1+x)/x]'/x^(k-3/2)（lim(x→0⁺)ln(1+x)/x=1）
=1/(1-2k)*lim(x→0⁺)[ln(1+x)/x]'/x^(k-3/2)
ln(1+x)=Σ(n:1→∞)(-1)^(n-1)*x^n/n=x-x^2/2+x^3/3-x^4/4...
ln(1+x)/x=Σ(n:1→∞)(-1)^(n-1)*x^(n-1)/n=1-x/2+x^2/3-x^3/4...
[ln(1+x)/x]'=Σ(n:2→∞)(-1)^(n-1)*(n-1)*x^(n-2)/n=-1/2+2x/3-3x^2/4...
原式=1/(1-2k)*lim(x→0⁺)[ln(1+x)/x]'/x^(k-3/2)
=1/(1-2k)*lim(x→0⁺)(-1/2+2x/3-3x^2/4...)/x^(k-3/2)=C
∴k-3/2=0,k=3/2
C=1/(1-2k)*(-1/2)=1/(1-3)*(-1/2)=1/4