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

Answer 1

#0#

#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#

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Answer 2

Aurora Alvarado

To find the limit of cos(t)/t^2 as t approaches infinity, we can use the concept of limits. By applying L'Hôpital's rule, we differentiate the numerator and denominator separately with respect to t.

Differentiating the numerator, we get -sin(t), and differentiating the denominator, we get 2t.

Taking the limit as t approaches infinity, we have:

lim (t->oo) cos(t)/t^2 = lim (t->oo) -sin(t)/(2t)

Since the limit of -sin(t)/(2t) as t approaches infinity is 0, the limit of cos(t)/t^2 as t approaches infinity is also 0.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you find #lim cost/t^2# as #t-&gt;oo#?

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