Question

What is the limit of `x^(1/x)` as `x->oo`?

Answer 1

`Lt_(x->oo)x^(1/x)=1`

Explanation:

Let `x^(1/x)=e^a`, then taking natural log on both sides, we get

`1/xlnx=a` and `x^(1/x)=e^(lnx/x)`

Hence, `Lt_(x->oo)x^(1/x)=Lt_(x->oo)e^(lnx/x)`

and as `e^u` is continuous `Lt_(x->oo)x^(1/x)=e^(Lt_(x->oo)(lnx/x))`

Now in `Lt_(x->oo)e^(lnx/x)` as `x->oo`, both numerator and denominators tend to infinity,

therefore we can apply L'Hospital's Rule and

`Lt_(x->oo)(lnx/x)=Lt_(x->oo)((1/x)/1)=Lt_(x->oo)1/x=0`

Hence `Lt_(x->oo)x^(1/x)=e^0=1`