Question

What is `\lim _ { x \rightarrow 1^ { + } } \frac { x } { x - 1} - \frac { 1} { \ln ( x ) }`?

Answer 1

`1/2`

Explanation:

Writing
`(xln(x)-(x-1))/((x-1)*ln(x))`
Using the rules of L'Hosptal we get
`lim_(x to 1^+)(ln(x)/(ln(x)+(x-1)/x))`
and this is
`lim_{x to 1^+)((x*ln(x))/(xln(x)+x-1))`
`lim_(x to 1^+)(1+ln(x))/(ln(x)+2)=1/2`