Question

Use the Squeeze Theorem to show that lim(t→[infinity]) cos⁵t/5t = 0. Complete the following argument: For t arbitrarily large and positive,

a) The numerator is always less than or equal to 1.
b) The denominator approaches infinity.
c) The limit of cos⁵t as t approaches infinity is 1.
d) The squeeze theorem is not applicable in this case.

Answer 1

To show that lim(t→[∞]) cos⁵t/5t = 0, we use the Squeeze Theorem which confirms that as t grows, the values of -1/5t and 1/5t both squeeze the function cos⁵t/5t towards 0.

Step-by-step explanation:

We are tasked with showing that lim(t→[∞]) cos⁵t/5t = 0 using the Squeeze Theorem. Let us examine the argument given:

1. The numerator cos⁵t is always less than or equal to 1 for all values of t since the maximum value of cos(t) is 1 and raising it to any power will not exceed 1.
2. The denominator 5t approaches infinity as t grows. Because we are dividing by an increasingly large number, our fraction will become smaller.
3. The statement that the limit of cos⁵t as t approaches infinity is 1 is incorrect. The cost function oscillates between -1 and 1, so it does not have a limit as t approaches infinity.
4. The use of the Squeeze Theorem is indeed applicable because it can help in showing the behavior of the given function between two other functions that have known limits at infinity.

To utilize the Squeeze Theorem, we need two other functions that 'squeeze' our function as t grows. Since -1 ≤ cos(t) ≤ 1, raising these to the fifth power we get -1 ≤ cos⁵(t) ≤ 1. Dividing by 5t, we have -1/5t ≤ cos⁵(t)/5t ≤ 1/5t. As t approaches infinity, both -1/5t and 1/5t approach 0. Therefore, by the Squeeze Theorem, the limit of cos⁵(t)/5t as t approaches infinity also equals 0.