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

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Answer 1 · 2017-12-13T21:43:38

Look below

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.

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

$lim_(x->oo) x$ = $oo$

Answer 2 · 2017-12-13T22:02:23

$e$

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

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

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