# Verify if the value of the limit of the function f(x)=(x^2-6x+5)/(x-1) is 4 x-->1

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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.**

First, we'll verify if the limit exists, for x = 1.

We'll substitute x by 1 in the expression of the function.

lim f(x) = lim (x^2-6x+5)/(x-1)

lim (x^2-6x+5)/(x-1) = (1-6+5)/(1-1) = 0/0

We notice that we've get an indetermination case.

We could apply 2 methods for solving the problem.

The first method is to calculate the roots of the numerator. Since x = 1 has cancelled the numerator, then x = 1 is one of it's 2 roots.

We'll use Viete's relations to determine the other root.

x1 + x2 = -(-6)/1

1 + x2 = 6

x2 = 6 - 1

x2 = 5

We'll re-write the numerator as a product of linear factors:

x^2-6x+5 = (x-1)(x-5)

We'll re-write the limit:

lim (x - 1)(x - 5)/(x - 1)

We'll simplify:

lim (x - 1)(x - 5)/(x - 1) = lim (x - 5)

We'll substitute x by 1:

lim (x - 5) = 1-5

lim (x^2-6x+5)/(x-1) = -4

**So, the limit of the given function is -4 and not 4. lim (x^2-6x+5)/(x-1) = -4, for x->1.**

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