#lim x->oo# #e^x((2^(x^n))^(1/e^x)-(3^(x^n))^(1/e^x))/x^n# , n belongs to N, is equal to ?

1 Answer
Cesareo R.
Jun 24, 2017

# log_e(2/3)#

Explanation:

Calling #u(x) = e^x# and #v(x)= x^n# we have

#e^x((2^(x^n))^(1/e^x)-(3^(x^n))^(1/e^x))/x^n equiv (u(x))/(v(x))(2^( (v(x))/(u(x)))-3^( (v(x))/(u(x))))# and now making #(u(x))/(v(x))=xi(x)# we have

#lim_(x->oo)e^x((2^(x^n))^(1/e^x)-(3^(x^n))^(1/e^x))/x^n equiv lim_(x->oo)xi(x)(2^(1/(xi(x)))-3^(1/(xi(x))))#

but #xi(x)->oo# as #x->oo# so

#lim_(x->oo)e^x((2^(x^n))^(1/e^x)-(3^(x^n))^(1/e^x))/x^n equiv lim_(y->oo)((2^(1/y)-3^(1/y))/(1/y)) = log_e(2/3)#

NOTE:

#lim_(y->oo)((2^(1/y)-3^(1/y))/(1/y)) equiv lim_(z->0)(2^z-3^z)/z = log_e(2/3)#

because defining #f(z) = 2^z-3^z# we have

#lim_(h->0)(f(0+h)-f(0))/h=lim_(h->0)((2^h-3^h)-(1-1))/h = f'(0)# and

#f'(z) = 2^z log_e 2-3^z log_e 3# and then

#f'(0)=log_e(2/3)#

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