Question

How do you find the limit of x^(1/lnx) as x approach 0 from positive side?

x

Answer 1

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

Explanation:

Let us first plug in `0` right away, keeping in mind that we're approaching from the right side.

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

This is an indeterminate form, so we must use l'Hospital's rule.

This function does not appear to be in rational function form, `f(x)/g(x)`; however, a little manipulation will yield that form.

In fact, the following route is usually the one that should be taken when dealing with the indeterminate limit of a function to the power of a function:

Let

`y=x^(1/lnx)`

Then,

`lny=ln(x^(1/lnx))` (Applying the logarithm to both sides)

`lny=1/lnx(lnx)=ln(x)/ln(x)` As `ln(x^a)=aln(x)`

Simplifying yields

`lny=1`

`lim_(x->0^+)lny=lim_(x->0^+)1=1`

So, we have the limit for `lny,` but we really want the limit for `y.` This is no issue, as

`lim_(x->0^+)y=lim_(x->0^+)e^lny=e^1=e`

So, `lim_(x->0^+)x^(1/lnx)=e`