How do you find the limit of #(1+7/x)^(x/10)# as x approaches infinity?

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Answer 1 · 2016-04-28T14:14:47

There are other method available, but I use #lim_(mrarroo)(1+1/m)^m = e#

#(1+7/x)^(x/10) = ((1+7/x)^x)^(1/10) = [(1+1/(x/7))^(x/7)]^(7/10)#

As #xrarroo# the quotient #x/7 rarroo# as well.

So, with #m=x/7# we see that the expression in the brackets goes to #e#. That is

#lim_(mrarroo)[(1+1/(x/7))^(x/7)] = e#

Therefore,

#lim_(xrarroo)(1+7/x)^(x/10) = lim_(xrarroo)[(1+1/(x/7))^(x/7)]^(7/10)#

# = [lim_(xrarroo)(1+1/(x/7))^(x/7)]^(7/10)#

# = e^(7/10)#