Question

Evaluate the limit using Squeeze/Sandwich Theorem:
$\displaystyle \large{ \lim_{n \to \infty} \frac{1}{n} \cos \frac{n \pi}{3} }$
Please show your work too — thanks!​

Answer 1

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

$-1 \le \cos\left(\dfrac{n\pi}3\right) \le 1 \implies -\dfrac1n \le \dfrac1n\cos\left(\dfrac{n\pi}3\right) \le \dfrac1n$

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

Answer:

$\displaystyle \lim_{n\rightarrow \infty}\left[\frac{1}{n}\cos\left(\frac{n\pi}{3}\right)\right]=0$

Step-by-step explanation:

The squeeze theorem states that if $g(x)\leq f(x)\leq h(x)$ for all $x\neq c$ 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 $\displaystyle \frac{1}{n}(1)$ will always be greater than or equal to  $\displaystyle \frac{1}{n}\cos\left( \frac{n\pi}{3}\right)$ and the function $\displaystyle \frac{1}{n}(-1)$ will always be less than or equal to $\displaystyle \frac{1}{n}\cos \left(\frac{n\pi}{3}\right)$.

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

We can easily compute $\displaystyle \lim_{n\rightarrow \infty}\left[- \frac{1}{n}\right]$ and $\displaystyle \lim_{n\rightarrow \infty}\left[\frac{1}{n}\right]$ with direct substitution.

Therefore, we have:

$\displaystyle \lim_{n\rightarrow \infty}\left[- \frac{1}{n}\right]=-\frac{1}{\infty}=0\\\\\displaystyle \lim_{n\rightarrow \infty}\left[ \frac{1}{n}\right]=\frac{1}{\infty}=0$

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