What the limit of (1/x)^1/x as x approaches oo? help me please

1 Answer
Dec 30, 2017

lim_(xrarr+oo)(1/x)^(1/x)=1

Explanation:

lim_(xrarr+oo)(1/x)^(1/x) = ?

(1/x)^(1/x) = e^(ln(1/x)^(1/x)) =e^(1/xln(1/x)) = e^(1/x(ln1-lnx) = e^(-lnx/x) = 1/e^(lnx/x)

so lim_(xrarr+oo)(1/x)^(1/x) =

lim_(xrarr+oo)e^(1/xln(1/x) =

lim_(xrarr+oo)1/e^(lnx/x) =A

  • lnx/x=u

x->+oo

u->0

because lim_(xrarr+oo)(lnx/x)=lim_(xrarr+oo)1/x=0
, lim_(xrarr+oo)lnx=+oo , lim_(xrarr+oo)x=+oo

y=lnx
graph{lnx [-10, 10, -5, 5]}

(using Rules De L'Hospital)

As a result, A= lim_(xrarr+oo)1/e^(lnx/x)

=lim_(urarr0)1/e^u =1/e^0=1