First slide
Introduction to limits
Question

The value of limx0(1+x)1/xex, is 

Moderate
Solution

We have,

(1+x)1/x=elog(1+x)x=e1xxx22+x33 (1+x)1/x=e1x2+x23=eex2+x23limx0(1+x)1/xex=limx0eex2+x23ex

=elimx0ex/2+x2/3+1x2+x23+×x2+x2x+x=12 e 

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