#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#