What is #lim_(x->0) ((1+x)^n-1)/x# ?

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Answer 1 · 2017-11-04T11:59:04

#lim_(x->0) ((1+x)^n-1)/x =n#

If we assume the question was: #lim_(x->0) ((1+x)^n-1)/x #

We have an indeterminate limit of the form #0/0#

Hence, L'Hopital's rule applies.

#:.lim_(x->0) ((1+x)^n-1)/x = lim_(x->0) (d/dx((1+x)^n-1))/(d/dx(x))#

#= lim_(x->0) (n(1+x)^(n-1)xx1 -0)/1# [Power rule and chain rule]

#= lim_(x->0) n(1+x)^(n-1)#

#= nxx1^(n-1)= n#