# `lim_(x->1^+) (x/(x - 1) - 1/ln(x))` Evaluate the limit.

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### 2 Answers

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

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

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

Apply L'Hospital's rule, Test L'Hospital condition :0/0

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

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

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

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

Again apply L'Hospital's rule, Test L'Hospital condition:0/0

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

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

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

Now plug in the value and simplify,

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

`=1/2`

`lim_(x->1^+) (x/(x-1)-1/lnx)`

To evaluate the limit of this function as x approaches 1 from the right, assign values to x that is greater than 1, but very near 1. Then, plug-in that value to the function

`f(x) = x/(x-1) - 1/lnx`

Let x = 1.01

`f(1.01) =1.01/(1.01-1)-1/ln1.01=0.5008291928`

Let x=1.001

`f(1.001)=1.001/(1.001-1)-1/ln1.001=0.5000832906`

Let x = 1.0001

`f(1.0001)=1.0001/(1.0001-1)-1/ln1.0001=0.5000083162`

Notice that as the value of x is getting closer and closer to 1, the value of f(x) approaches 0.5 .

**Therefore, `lim_(x->1^+) (x/(x-1)-1/lnx)=1/2 .` **