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The limit `lim_(x-> 1)(x^2-7x+6)/(x-1)` has to be determined.
If we substitute x with 1, x^2 - 7x + 6 = 1 - 7 + 6 = 0 and x - 1 = 1 - 1 = 0.
The expression `(x^2-7x+6)/(x-1) = 0/0` which is indeterminate. As a result we can use l'Hospital's rule and replace the numerator and denominator by their derivatives.
`(x^2-7x+6)' = 2x - 7`
`(x-1)' = 1`
The limit is now:
`lim_(x->1) (2x - 7)/1`
Substituting x = 1 gives `(2*1-7)/1 = -5`
The required limit `lim_(x-> 1)(x^2-7x+6)/(x-1) = -5`
First, we'll verify if the limit exists, for x = 1, so, we'll substitute x by 1 in the expression of the function.
lim f(x) = lim (x^2-7x+6)/(x-1)
lim (x^2-7x+6)/(x-1) = (1-7+6)/(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 = -(-7)/1
1 + x2 = 7
x2 = 7 - 1
x2 = 6
We'll re-write the numerator as a product of linear factors:
x^2-7x+6 = (x-1)(x-6)
We'll re-write the limit:
lim (x-1)(x-6)/(x - 1)
We'll divide by (x-1):
lim (x-1)(x-6)/(x - 1) = lim (x - 6)
We'll substitute x by 1:
lim (x - 6) = 1-6
lim (x - 6) = -5
The limit of the given function f(x) = (x^2-7x+6)/(x-1) is:
lim (x^2-7x+6)/(x-1) = -5
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