impossible limit problem

[zeehan]

need help with this limit problem----> lim x->2 (t^3+3t^2-12t+4)/t^3-4t

this one is from the book calculus-early transcendentals by Anton Bivens, 9th edition, exercise set 1.2 of limits ( problem no.13)

stuck with the factorization....please help...

 

[HallsofIvy]

I presume you mean lim t->2 (t^3+3t^2-12t+4)/(t^3-4t).

The obvious first thing to do is set t= 2: (8- 12- 24+ 4)/(8- 8)= 0/0 which is "inconclusive".

The fact that each of numerator and denominator is 0 when t= 2 means that t- 2 is a factor.

Divide each of t^3+ 3t^2- 12t+ 4 and t^3- 4t by t- 2 to get the other factor. ("Synthetic division" was specifically developed for that job!)

Or you could note that t^3- 4t= t(t^2- 4) (and t^2- 4 is easy to factor) and that t^3- 3t^2- 12t- 4= t^3- 2t^2- 10t- 2t- 4 = t(t^2- 3t- 10)- 2(t- 2).

 

[zeehan]

This was not my problem---> t^3- 3t^2- 12t- 4

it was , t^3+3t^2-12t+4

you made a mistake with the signs..but this technique works..thanks a lot for the help...

 

[zeehan]

another similar one but tougher---> lim t-->1 (t^3+t^2-5t+3)/(t^3-3t+2)

the denominator is the main problem....

 

[HallsofIvy]

Once again, the only reason that is "tough" (and certainly NOT "impossible") is that if you set t= 1 in both numerator and denominator, you get 1+1- 5+ 3= 0 and 1- 3+ 2= 0. And that tells you that each has t- 1 as a factor.

So divide each by t- 1 to see what the other factor is:
$\displaystyle \frac{t^3+ t^2- 5t+ 3}{t- 1}= t^2+ 2t- 3$
$\displaystyle \frac{t^2- 3t+ 2}{t- 1}= t- 2$