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

2 Answers
Jan 3, 2018

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

Jan 3, 2018

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