Question

What is the limit using L' Hopital's Rule?

`lim_(x->0) ln(x)/x`

Answer 1

See below.

Explanation:

L'Hopital's Rule will give us:

`d/dx(ln(x))=1/x`

`d/dx(x)=1`

This gives us `1/x`

We no longer have an indeterminate form so we can't use the rule again.

as `x ->0^-` , `1/x->-oo`

as `x ->0^+` , `1/x->oo`

So:

`lim_(x->0)(ln(x))/x=`undefined

Answer 2

Because `lnx` is imaginary for `x < 0`, I assume that we want `lim_(xrarr0^+) lnx/x`

Explanation:

`lim_(xrarr0^+) lnx = -oo` and `lim_(xrarr0^+)x = 0^+`

Therefore `lim_(xrarr0^+) lnx/x` has the form `(-oo)/0^+`

No use of l'Hopitals' rule is needed or permitted.

`lim_(xrarr0^+) lnx/x = -oo`

If you try to use l'Hopital, you get an incorrect answer.

`lim_(xrarr0^+)((1/x)/1) = oo` which is not correct.

Here is the graph of `y=lnx/x`

graph{(lnx)/x [-9.21, 13.29, -7.425, 3.825]}