How do you find #lim cost/t^2# as #t->oo#?

1 Answer
May 10, 2017

#0#

Explanation:

#cost# oscillates between the values of #-1# and #1#.

The denominator, #t^2#, approaches #oo# as #trarroo#.

It's fairly simple to see that no matter what the value of #cost# is, it will be significantly "overpowered" by the growth of #t^2# in the denominator.

As #t# gets sufficiently large, we will have very large values in the denominator and only value between #-1# and #1# in the numerator.

Thus, we get values like #1/10000# and #-1/100000000# as #t# increases. These terms get closer and closer to being #0#.

So:

#lim_(trarroo)cost/t^2=0#