Question

How do you find the limit of `[ 1/ln(x) - 1/(x-1) ]` as x approaches 1?

Answer 1

Combine, using a common denominator, then use L'Hôpital's rule , twice.

Explanation:

`lim_(xto1)[1/ln(x) - 1/(x - 1)] = ?`

Make a common denominator:

`lim_(xto1)[1/ln(x)(x - 1)/(x - 1) - 1/(x - 1)ln(x)/ln(x)] =`

`lim_(xto1)[(x - 1)/(ln(x)(x - 1)) - ln(x)/(ln(x)(x - 1))] =`

`lim_(xto1)[(x - 1 - ln(x))/(ln(x)(x - 1))]`

The above evaluates at the limit to an indeterminate form, `0/0`. Therefore, the use of L'Hôpital's rule is warranted:

Compute the first derivative of the numerator:

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

Compute the first derivative of the denominator:

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

Make a new fraction out of the new numerator and new denominator:

`lim_(xto1)[(1 -1/x)/((x - 1)/x + ln(x))]`

Multiply by `x/x`

`lim_(xto1)[(x -1)/(x - 1 + xln(x))]`

It is still the indeterminate form `0/0`, therefore, we apply the rule, again:

Numerator:

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

Denominator:

`(d(x - 1 + xln(x)))/dx = 1 + ln(x) + x/x = 2 + ln(x)`

Here is the new expression:

`lim_(xto1)[1/(2 + ln(x))]`

The above can be evaluated at the limit:

`1/(2 + ln(1)) = 1/2`

Therefore, the original expression has the same limit:

`lim_(xto1)[1/ln(x) - 1/(x - 1)] = 1/2`