Question

Evaluate the limit using Squeeze/Sandwich Theorem:


`\displaystyle \large{ \lim_(n \to \infty) (1)/(n) \cos (n \pi)/(3) }`
Please show your work too — thanks!​

Answer 1

cos(x) is bounded between -1 and 1. So,

`-1 \le \cos\left(\frac{n\pi}3\right) \le 1 \implies -\frac1n \le \frac1n\cos\left(\frac{n\pi}3\right) \le \frac1n`

As n goes to infinity, both -1/n and 1/n converge to 0, so

`\displaystyle\lim_(n\to\infty)\frac1n\cos\left(\frac{n\pi}3\right) = \boxed{0}`

as well.

Answer 2

`\displaystyle \lim_(n\rightarrow \infty)\left[(1)/(n)\cos\left((n\pi)/(3)\right)\right]=0`

Explanation:

The squeeze theorem states that if
`g(x)\leq f(x)\leq h(x)` for all
`x\\eq c` in some interval around
`c` and
`\displaystyle \lim_(x\rightarrow c)g(x)=L` and
`\displaystyle \lim_(x\rightarrow c)h(x)=L`, then it follows that
`\displaystyle \lim_(x\rightarrow c)f(x)=L`.

Essentially what this is saying is that if a function
`g` is always smaller than or equal to function
`f` in some interval and a function
`h` is always greater than or equal to
`f` in the same interval, if
`g` and
`h` approach the same value at some point,
`f` must also approach that point, since it is being "squeezed", hence the name squeeze theorem.

Recall that the maximum output of cosine is 1 and the minimum output of cosine is -1. A quick check on the unit circle will confirm this.

Therefore, the function
`\displaystyle (1)/(n)(1)` will always be greater than or equal to
`\displaystyle (1)/(n)\cos\left( (n\pi)/(3)\right)` and the function
`\displaystyle (1)/(n)(-1)` will always be less than or equal to
`\displaystyle (1)/(n)\cos \left((n\pi)/(3)\right)`.

Hence,
`\displaystyle -(1)/(n)\leq (1)/(n)\cos\left( (n\pi)/(3)\right)\leq (1)/(n)\text{ for all } x\in \mathbb{R}`.

We can easily compute
`\displaystyle \lim_(n\rightarrow \infty)\left[- (1)/(n)\right]` and
`\displaystyle \lim_(n\rightarrow \infty)\left[(1)/(n)\right]` with direct substitution.

Therefore, we have:

`\displaystyle \lim_(n\rightarrow \infty)\left[- (1)/(n)\right]=-(1)/(\infty)=0\\\\\displaystyle \lim_(n\rightarrow \infty)\left[ (1)/(n)\right]=(1)/(\infty)=0`

Since
`\displaystyle \lim_(n\rightarrow \infty)\left[- (1)/(n)\right]=\displaystyle \lim_(n\rightarrow \infty)\left[ (1)/(n)\right]=0` and
`\displaystyle -(1)/(n)\leq (1)/(n)\cos\left( (n\pi)/(3)\right)\leq (1)/(n)`, then from the squeeze theorem,
`\displaystyle \lim_(n\rightarrow \infty)\left[(1)/(n)\cos\left((n\pi)/(3)\right)\right]= \lim_(n\rightarrow \infty)\left[- (1)/(n)\right]=\displaystyle \lim_(n\rightarrow \infty)\left[ (1)/(n)\right]=\boxed{0}`