Lim(e^x+x)^(1/x) as x→0+?

1 Answer
Jim S
May 24, 2018

#lim_(x->0^+)(e^x+x)^(1/x)=e^2#

Explanation:

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

  • #(e^x+x)^(1/x)=e^(ln(e^x+x)^(1/x))=e^(ln(e^x+x)/x)#

#lim_(x->0^+)ln(e^x+x)/x=_(DLH)^((0/0))##lim_(x->0^+)((ln(e^x+x))')/((x)')# #=#

#lim_(x->0^+)(e^x+1)/(e^x+x)=2#

Therefore,

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

Set

#ln(e^x+x)/x=u#
#x->0^+#
#u->2#

#=# #lim_(u->2)e^u=e^2#

Related questions
Impact of this question
1957 views around the world
You can reuse this answer
Creative Commons License