Question

What is `lim_(x=>0^+)x^(1/x)` ?

`lim_(x=>0^+)x^(1/x)`

Answer 1

`0`

Explanation:

Let `L = lim_(x to 0^+)x^(1/x)`

`ln L = ln (lim_(x to 0^+)x^(1/x))`

Because `ln x` is continuous for `x > 0` it follows that:
`ln L = lim_(x to 0^+) ln (x^(1/x))`

`implies ln L = lim_(x to 0^+) (ln x)/x`

By the product rule:

`lim_(x to 0^+) (ln x)/x = lim_(x to 0^+) ln x * lim_(x to 0^+) (1)/x`

And

• `lim_(x to 0^+) (ln x) = - oo`

• `lim_(x to 0^+) (1)/x = oo`

Thus:

`ln L = - oo`

`implies L = lim_(x to 0^+)x^(1/x) = e^(-oo) = 0`

Answer 2

Set `f(x)=x^(1/x)` then take logarithms on both sides

`lnf(x)=1/x*lnx`

`f(x)=e^(lnx/x)`

Now we must find the limit `lim_(x->0^+) lnx/x` .

We observe that this is `lim_(x->0^+) lnx/x=(-oo)/0^+`

which is actually "equal" to negative infinity .
(A very big negative number which is `-oo` divided by a small number which is `0^+`)

Hence the final limit is

`lim_(x->0^+) f(x)=e^(-oo)=0`