How do you find the limit of #(1+1/x)^x# as x approaches infinity?

1 Answer
Jim H
Oct 16, 2016

#lim_(urarroo)(1+1/u)^u = e# is sometimes taken as the definition of #e#.

It can be verified using l'Hospital's Rule

#ln((1+1/u)^u) = u ln(1+1/u) = ln(1+1/u)/(1/u)# which takes indeterminate form #(-oo)/oo# as #xrarroo#

L'Hospitals's rule asks us to eveluate

#lim_(urarroo)(1/(1+1/u)(-1/u^2))/(-1/u^2)# which evaluates to #1#.

Since the limit of the #ln# is #1#, the limit of the expression is #e#.

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