I need help finding the limit of this problem?

Question

The limit as t approaches 2 of:
(t^3 + 3t^2 -12t +4)/(t^3-4t)

When I plug in 2 I get 0 over 0. I know this needs to be factored. If it helps, I know the answer is 3/2.

Answer 1

Since you know that its 0 over 0 after substituting 2, you are sure that the numerator and denominator have a factor (t-2) which needs to be cancelled out.

So, dividing (by long or synthetic division) the numerator by t-2 gives t^2+5t-2. The denominator is easily factored t(t-2)(t+2).

Thus, after factoring and cancelling you get

LIMIT (t^2+5t-2)/(t^2+2t)
t->2

you can do now your favorite substitution technique. :)

Answer 2

We are doing limits too!

You are right that it needs to be factored. Hm. Let's see...

I'm using synthetic division to find a binomial factor of the cubic numerator. When I do that, I get this:

[(x - 2)(x^2 + 5x - 2)] / [t(t^2 - 4)]

Factor further.

[(t - 2)(t^2 + 5t - 2)] / [t(t - 2)(t + 2)]

This means that the (t - 2)'s cancel each other.

limit as t approaches 2 of:
(t^2 + 5t - 2) / t(t + 2)

Plug in 2.

(2^2 + 5(2) - 2) / 2(2 + 2)

(4 + 10 - 2) / 2(4)

(14 - 2) / 8

12/8

3/2

Answer 3

(t^3 + 3t^2 - 12t +4)/(t^3-4t) = (t^3 + 3t^2 - 12t +4)/{t(t-2)(t+2)}

Try to factor numerator assuming (t-2) is one of the factors:

(t-2)(t^2+at+b) = t^3 + (a-2)T^2 + (b-2a)t - 2b

Now comparing to above numerator

a-2 = 3

b-2a = -12

-2b =4 -----> b = -2 so a = 5

Then above equation reduces to

(t^2+5t -2)/(t(t+2)) ---> t->2 (4+10-2)/(2*(4)) = 12/8 =3/2