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.
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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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