lim_(x->oo) x^(1/log_e x) = ?

2 Answers
Dec 13, 2017

Look below

Explanation:

since we know that ln(oo) = oo, we can state that lim_(x->oo) oo/ln(oo)=oo/oo, which is indeterminate form.

indeterminate form is 0/0 or oo/oo

now we use l'hopital's rule by taking the derivative of both sides.

\frac{\frac(d}{dx} [x]}{\frac{d}{dx} [ln(x)] = \frac{1}{1/x} = x/1 = x

lim_(x->oo) x = oo

Dec 13, 2017

e

Explanation:

Calling y = x^(1/log_e x) we have

log_e y = 1/(log_e x)log_e x = 1 then

y = x^(1/log_e x) = e

then

lim_(x->oo) x^(1/log_e x) = e