# `lim_(x->oo) x^(1/x)` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply, explain...

`lim_(x->oo) x^(1/x)` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply, explain why.

*print*Print*list*Cite

Replacing `oo` for x in limit equation yields the nedetermination `oo^o` . You need to use the logarithm technique, such that:

`f(x) = (x)^(1/x)`

You need to take logarithms both sides, such that:

`ln f(x) = ln (x)^(1/x))`

Using the property of logarithms yields:

`ln f(x) = (1/x)*ln x`

`ln f(x) = (ln x)/x`

You need to evaluate the limit:

`lim_(x->oo) ln f(x) = lim_(x->oo) (ln x)/x = oo/oo`

You need to use L'Hospital theorem:

`lim_(x->oo) (ln x)/x = lim_(x->oo) ((ln x)')/(x') `

`lim_(x->oo) ((ln x)')/(x') = lim_(x->oo) (1/x)/1`

`lim_(x->oo) 1/x = 1/oo = 0`

Hence, `lim_(x->oo) ln f(x) = 0` , such that `lim_(x->oo) f(x) = e^0 = 1`

**Hence, evaluating the given limit, using l'Hospital rule and logarithm technique yields `lim_(x->oo) (x)^(1/x) = e^0 = 1` .**

Plug in `oo ` everywhere you x

`lim_(x-gtoo)x^(1/oo)`

1 divided by a really big number is just 0

Therefore,

`lim_(x-gtoo)x^0`

`=1`