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

Question

1058 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-30T19:26:53

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

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

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

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

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

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

$x->+oo$

$u->0$

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

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

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