How to solve this problem #lim(e^x-x)^(1/x))# when x --> 0 ?

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Answer 1 · 2018-01-05T14:04:32

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

#lim_(xrarr0)(e^x-x)^(1/x)#

#(e^x-x)^(1/x)=e^(ln(e^x-x)^(1/x))# #=# #e^(1/xln(e^x-x))#

So now we need to calculate
#lim_(xrarr0)ln(e^x-x)/x#

The limit is #0/0# so we can use De L'Hospital Rules :

#lim_(xrarr0)ln(e^x-x)/x# #=lim_(xrarr0)(((e^x-x)')/(e^x-x))/1# #=#

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

Therefore,

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

Set #ln(e^x-x)/x=u#

#x->0#
#u->0#

#=# #lim_(urarr0)e^u=e^0=1#