How do you find the limit of # (e^x-1)^(1/lnx)# as x approaches 0?

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Answer 1 · 2016-06-16T08:02:15

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

To avoid any complications in domain with regard to #ln(x)#, we will take the limit as #x->0^+#

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

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

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

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

(Because #e^x# is a continuous function, we have #lime^f(x) = e^(limf(x))#)

Then, it remains to evaluate #lim_(x->0^+)ln(e^x-1)/ln(x)#

As #lim_(x->0^+)ln(x)=-oo#, direct substitution leads to an #oo/oo# indeterminate form. Thus, we can apply L'Hopital's rule.

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

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

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

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

(As this leads to a #0/0# indeterminate form, we apply L'Hopital's rule once more.)

#=lim_(x->0^+)(d/dx(xe^x))/(d/dx(e^x-1))#

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

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

#=1#

Substituting back into the original problem, we find our answer:

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

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

#=e^1#

#=e#