Evaluate the limit of the function (x^2-3x+2)/(x^2-4) using L'Hospital rule.

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.