Friday, 29 May 2015

real analysis - Evaluating limntoinftynleft(1frac1eleft(1+frac1nright)nright)





limnn(11e(1+1n)n)




If I write expansion of (1+1n)n it was equal to expansion of e so nn=0. Is limit is zero ?



Edit:
Uploading screen shot of my response. They marked it correct and I don't know if answer key is wrong or not ? Please can anyone say for sure like with 100 percent surety if answer given is wrong. enter image description here


Answer




By the L'Hôpital's rule we obtain:
limn+n(11e(1+1n)n)=limx0+e(1+x)1xex=1elimx0+(eln(1+x)x)=


=1elimx0+(1+x)1x(1x(1+x)ln(1+x)x2)=

=1elimx0+(1+x)1xlimx0+x(1+x)ln(1+x)x2+x3=

=limx0+1ln(1+x)12x+3x2=limx0+ln(1+x)xlimx0+12+3x=12.


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