Since you did not specified the accumulation point, we'll
consider 2 cases.
In the 1st case, the accumulation point
is infinite.
We'll calculate the limit by factorizing both,
numerator and denominator, by x^2:
lim
(x^2-5x+6)/(x^2+4x-21) = lim
x^2(1-5/x+6/x^2)/x^2(1+4/x-21/x^2)
We'll
simplify:
lim (1-5/x+6/x^2)/(1+4/x-21/x^2) = 1/1 =
1
lim (x^2-5x+6)/(x^2+4x-21) =
1
In the 2nd case, the accumulation point is
3.
To calculate the limit, we'll substitute x by
3:
lim (x^2-5x+6)/(x^2+4x-21) =
(9-15+6)/(9+12-21)
We notice that we've get an
indeterminacy: 0/0
Since x = 3 cancells both numerator and
denominator, that means that x = 3 is one of the roots.
lim
(x^2-5x+6)/(x^2+4x-21) = lim (x-3)(x-2)/(x-3)(x+7)
We'll
simplify and we'll get:
lim (x^2-5x+6)/(x^2+4x-21) = lim
(x-2)/(x+7)
To calculate the limit, we'll substitute x by
3:
lim (x-2)/(x+7) =
(3-2)/(3+7)
lim (x-2)/(x+7) =
1/10
So, if x->infinite, the value of
limit is 1 and if x->3, the value of limit is
1/10.
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