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

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gsarora17 eNotes educator| Certified Educator




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





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




Now plug in the value and simplify,




lemjay eNotes educator| Certified Educator

`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


Let x = 1.0001


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 .`     


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