We see that we can't simply equate this to the sum of the limits of all 3 terms, since the first term shoots off to +oo, while the second term approaches -oo. Different terms are pulling us towards different infinities. So what can we do?
Find the highest order of t in the polynomial, and factor it out from the whole thing:
color(white)=lim_(t->"-"oo)(4t^2+6t+2)
=lim_(t->"-"oo)t^2(4+6/t+2/t^2)
Then, use "the limit of a product is the product of the limits":
=[lim_(t->"-"oo)t^2] times [lim_(t->"-"oo)(4+6/t+2/t^2)]
Note: this rule is acceptable as long as we don't have one limit approaching +-oo while the other one goes to 0.
The first limit will approach +oo. The second will approach 4. This is because dividing by larger and larger (negative) numbers makes the terms 6/t and 2/t^2 get closer to 0, meaning their contributions to the sum 4+6/t+2/t^2 will diminish as t grows.
So we end up with
lim_(t->"-"oo)(4t^2+6t+2)=[lim_(t->"-"oo)t^2] times [lim_(t->"-"oo)(4+6/t+2/t^2)]
color(white)(lim_(t->"-"oo)(4t^2+6t+2))=[oo] times [4]
color(white)(lim_(t->"-"oo)(4t^2+6t+2))=oo
as our answer.
Bonus:
Different terms may be pulling us toward different infinities, but the term with the highest power of t is going to have the strongest "pull" as t gets closer to -oo. For lim_(x->+-oo) of simple polynomials, we only need to keep the term with the highest power of x. In other words:
lim_(t->"-"oo)(4t^2+6t+2)=lim_(t->"-"oo)4t^2
color(white)(lim_(t->"-"oo)(4t^2+6t+2))=+oo.
(The polynomial 4t^2+6t+2 is a parabola that opens up, so this answer makes sense.)
