Evaluate the limit of the function (x^2-3x+2)/(x^2-4) using L'Hospital rule.
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As you have not specified what x is tending to, I take it as 2.
We have to find the value of lim x--> 2 [(x^2-3x+2)/(x^2-4)]
If we substitute x = 2, we get the indeterminate form 0/0. So we use L'Hopital's Rule and substitute the numerator and the denominator by their derivatives
=> lim x-->2 [(2x -3)/(2x)]
substitute x = 2
=> (4 - 3) / 4
=> 1/4
The required value of lim x--> 2 [(x^2-3x+2)/(x^2-4)] = (1/4)
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You did not mentioned the accumulation point. Though, L'Hospital rule is used only in case of indetermination, so the accumulation point can only be a common solution of denominator and numerator of the fraction.
We'll calculate the roots of the equations:
x^2-3x+2 = 0
x1 = [3 + sqrt(9 - 8)]/2
x1 = (3+1)/2
x1 = 2
x2 = 1
We'll calculate the roots of denominator:
x^2 - 4 = 0
Since it is a difference of squares, we'll re-write it as a product:
(x-2)(x+2) = 0
x - 2 = 0
x = 2
x + 2 = 0
x = -2
We notice that the common root for both is x = 2.
We'll conclude that the accumulation point is x = 2, so x->2
We'll evaluate the limit:
lim (x^2-3x+2)/(x^2-4) = (4 - 6 + 2)/(4 - 4) = 0/0
We'll apply L'Hospital rule:
lim (x^2-3x+2)/(x^2-4) = lim (x^2-3x+2)'/(x^2-4)'
lim (x^2-3x+2)'/(x^2-4)' = lim (2x - 3)/2x
We'll substitute x by 2:
lim (2x - 3)/2x = (4 - 3)/4
lim (2x - 3)/2x = 1/4
The limit of the function is: lim (x^2-3x+2)/(x^2-4) = 1/4.
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