Final answer:
To prove the limit of the function (2 - cos(x))/(x + 3) as x approaches infinity is 0 using the Squeeze Theorem, two bounding functions are established: 0 ≤ f(x) ≤ 1/(3) and 1 ≤ 2 - cos(x) ≤ 3. By applying the Squeeze Theorem, it can be concluded that the limit of the given function is 0 as x approaches infinity.
Step-by-step explanation:
To prove that the limit of the function (2 - cos(x))/(x + 3) as x approaches infinity is 0 using the Squeeze Theorem, we need to find two other functions that are greater than or equal to and less than or equal to the given function, and whose limits as x approaches infinity are both 0.
First, note that the cosine function is bounded between -1 and 1. Therefore, we can say that -1 ≤ cos(x) ≤ 1 for all x. By adding 2 to all parts of the inequality, we get 1 ≤ 2 - cos(x) ≤ 3.
Now, let's consider the function f(x) = (1)/(x + 3). Since the denominator is increasing as x approaches infinity, the value of the function is decreasing. This means that 0 ≤ f(x) ≤ 1/(3). Therefore, we have established two bounding functions: 0 ≤ f(x) ≤ 1/(3) and 1 ≤ 2 - cos(x) ≤ 3.
According to the Squeeze Theorem, if two functions g(x) and h(x) both approach the same limit L as x approaches infinity, and if a third function f(x) is squeezed between the values of g(x) and h(x), then f(x) also approaches the limit L as x approaches infinity.
In this case, as x approaches infinity, both 0 ≤ f(x) = (2 - cos(x))/(x + 3) ≤ 1/(3) and 0 ≤ g(x) = 0 ≤ 1/(3) approach the limit 0. Therefore, the limit of the function (2 - cos(x))/(x + 3) as x approaches infinity is indeed 0.