Question #ab616

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Answer 1 · 2018-02-26T07:47:10

See the proof below

The MacLaurin Series is

#f(x)=f(0)+xf'(0)+x^2/(2!)f''(0)+x^3/(3!)f'''(0)+x^4/(4!)f^(iv)(0)+x^5/(5!)f^(v)(0)+x^5 epsilon()#

#f(x)=ln(1+x)#, #=>#, #f(0)=0#

#f'(x)=1/(1+x)#, #=>#, #f'(0)=1#

#f''(x)=-1/(1+x)^2#, #=>#, #f''(0)=-1#

#f'''(x)=2/(1+x)^3#, #=>#, #f'''(0)=2#

#f^(iv)(x)=-6/(1+x)^4#, #=>#, #f^(iv)(0)=-6#

#f^(v)(x)=24/(1+x)^5#, #=>#, #f^(v)(0)=24#

Therefore,

#ln(1+x)=x-x^2/2+x^3/3-x^4/4+x^5/5+x^5epsilon(x)#

#lim_(x->0)epsilon(x)=0#