Question

The limit of `ln(x)/(x-1)` as x approaches 1 equals what?

Answer 1

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

Explanation:

First, we can try directly pluggin in `x`:
`ln(1)/(1-1)=0/0`

However, the result `0 \/ 0` is inconclusive, so we need to use another method.

In this case, my method of choice would be L'Hôpital's rule. It says that you if you have a limit resulting in the indeterminate form `0/0`, you can differentiate both the numerator and the denominator, and if the limit of this exists, it will be the same as the original limit.

More formally, this can be expressed as:
If `lim_(x->c)f(x)=0` and `lim_(x->c)g(x)=0` and `lim_(x->c)(f'(x))/(g'(x))=L`

Then `lim_(x->c)(f(x))/(g(x))=L`

In our case, `f(x)=ln(x)` and `g(x)=x-1`. Both their derivatives are rather straight-forward:
`f'(x)=1/x`
`g'(x)=1`

This means we can rewrite our limit like so:
`lim_(x->1)ln(x)/(x-1)=lim_(x->1)(1/x)/1=lim_(x->1)1/x=1/1=1`

Answer 2

See below.

Explanation:

`lim_(x->1) lnx/(x-1) = lim_(x->1)ln(x^(1/(x-1))) = ln(lim_(x->1)x^(1/(x-1)))`

but making `y = x-1`

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

then

`lim_(x->1) lnx/(x-1) = ln(e) = 1`