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How do you solve the following limit #(e^x-1)/x# as x approaches zero?

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May 9, 2015

It is a remarkable limit, but, if you want to demonstrate it, you have to know the fundamental limit:

#lim_(xrarroo)(1+1/x)^x=e# (number of Neper), and also this limit:

#lim_(xrarr0)(1+x)^(1/x)=e# that it is easy to demonstrate in this way:

let #x=1/t#, so when #xrarr0# than #trarroo# and this limit becomes the first one.

So:

let #e^x-1=trArre^x=t+1rArrx=ln(t+1)#

and if #xrarr0rArrtrarr0#

#lim_(xrarr0)(e^x-1)/x=lim_(trarr0)t/ln(t+1)=lim_(trarr0)1/(ln(t+1)/t)=#

#=lim_(trarr0)1/(1/tln(t+1))=lim_(trarr0)1/(ln(t+1)^(1/t))=#

(for the second limit)#=1/(lne)=1/1=1#.

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mason m Share
Apr 9, 2017

The limit is in the indeterminate form #(e^0-1)/0=0/0#, so l'Hopital's rule applies:

#lim_(xrarr0)(e^x-1)/x=lim_(xrarr0)(d/dx(e^x-1))/(d/dx(x))=lim_(xrarr0)e^x/1=e^0=1#

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