Question

In Exercises 9-12, find the limit and confirm your answer using the Sandwich Theorem. $$\lim _{x \rightarrow \infty} \frac{1-\cos x}{x^{2}}$$

Short answer

The limit as x approaches infinity for the function \( \frac{1-\cos x}{x^{2}} \) is 0.

Step-by-step solution

Define sandwich functions

We need to set up two functions, one which must be always less than or equal to our original function and one which must be always greater than or equal to our function. The absolute version of the given function is \( \frac{1-\cos x}{x^{2}} \leq \frac{1+|\cos x|}{x^{2}}\). Thus we have \( \frac{0}{x^{2}} \leq \frac{1-\cos x}{x^{2}} \leq \frac{1+|\cos x|}{x^{2}}\), such that \( f(x) = 0 \), \(g(x) = \frac{1-\cos x}{x^{2}} \) and \(h(x) = \frac{1+|\cos x|}{x^{2}} \).

Compute the limit of f(x) and h(x)

We already know that \(f(x)\) equals 0 and thus its limit is 0. For \(h(x)\), we are dealing with \(\cos x\) which fluctuates between -1 and 1, thus \( |\cos x| \) fluctuates between 0 and 1. Thus, \(h(x)\) fluctuates between 0 and \(\frac{2}{x^{2}}\), the limit of which is 0 as \(x\) approaches infinity.

Apply the Sandwich Theorem

The Sandwich theorem states, in this case, that if the limit of \(f(x)\) equals the limit of \(h(x)\) as \(x\) approaches infinity, then the limit of \(g(x)\), our original function, must also be the same. We have shown that both limits are 0, therefore the limit of the original function is also 0.

Confirm the limit

To make sure the limit is indeed 0, it could also be computed directly using l'Hopital's rule. However, since the function is limit is proven mathematically to be 0, this step isn't necessary in this case.