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.)*