Find the limit via l'hopitals rule

lim X->1 [((x)/(x-1))-((1)/(Inx))]

Expert Answers

An illustration of the letter 'A' in a speech bubbles

You need to bring the fractions to a common denominator, such that:

`lim_(x->1) (x*lnx - x + 1)/((x - 1)*ln x)`

Substituting 1 for x yields:

`lim_(x->1) (x*lnx - x + 1)/((x - 1)*ln x) = (1*ln1 - 1 + 1)/((1 - 1)*ln1)`

`lim_(x->1) (x*lnx - x + 1)/((x - 1)*ln x) = 1*0/(0*0) = 0/0`

Since the limit is indeterminate, `0/0` , you may use l'Hospital's theorem, such that:

`lim_(x->1) ((x*lnx - x + 1)')/(((x - 1)*ln x)')`

You need to use the product rule such that:

`lim_(x->1) ((x*lnx - x + 1)')/(((x - 1)*ln x)') = lim_(x->1) (x'*lnx + x*(lnx)' - 1)/((x - 1)'*ln x + (x - 1)*(ln x)')`

`lim_(x->1) ((x*lnx - x + 1)')/(((x - 1)*ln x)') =lim_(x->1) (ln x + x*(1/x) - 1)/(ln x + (x - 1)/x)`

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

Substituting 1 for x yields:

`lim_(x->1) (ln x + 1 - 1)/(ln x + 1 - 1/x) = (ln 1 + 1 - 1)/(ln 1 + 1- 1/1) = 0/0`

You need to use l'Hospital's theorem again, such that:

`lim_(x->1) ((ln x + 1 - 1)')/((ln x + 1 - 1/x)') = lim_(x->1) (1/x)/(1 + 1/x^2)`

Substituting 1 for x yields:

`lim_(x->1) (1/x)/(1 + 1/x^2) = (1/1)/(1 + 1/1^2) = 1/2`

Hence, evaluating the given limit, using l'Hospital's theorem two times, yields `lim_(x->1) (x*lnx - x + 1)/((x - 1)*ln x) = 1/2.`

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial