We have to verify the value of lim x--> 1[ (x^2 - 6x + 5)/(x - 1)]

Substituting x = 1 we get the indeterminate form 0/0.

So, we can use l'Hopital's rule and substitute the numerator and denominator by their derivatives.

=> lim x--> 1[ (2x - 6)/1]

Substitute x = 1

=> 2 - 6

=> -4

**The value of lim x--> 1[ (x^2 - 6x + 5)/(x - 1)] is not 4 but -4.**