# Calculate the limit of the function (x^2-5x+6)/(x^2+4x-21).

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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.