Question

`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 ?

Answer 1

`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)`