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 -...

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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.` **