# How do you use the Squeeze Theorem to find lim ( sin (2pi/x)) as x approaches zero?

Oct 12, 2015

${\lim}_{x \rightarrow 0} \sin \left(\frac{2 \pi}{x}\right)$ Does Not Exist

#### Explanation:

Here is the key idea.
However close to $0$ we start, it will happen that as $x \rightarrow 0$,

we get infinitely many values of $\frac{2 \pi}{x}$ that are coterminal with $\frac{\pi}{2}$, so they have $\sin \left(\frac{2 \pi}{x}\right) = 1$. (Specifically, for every x=4/(1+4n with $n$ an integer.)

We also get infinitely many values of $\frac{2 \pi}{x}$ that are coterminal with $\frac{3 \pi}{2}$, so they have $\sin \left(\frac{2 \pi}{x}\right) = - 1$. (Specifically, for every x=4/(3+4n with $n$ an integer.)

$\sin \left(\frac{2 \pi}{x}\right)$ cannot be getting closer and closer to some number $L$, because we will always have some $x$'s for which the distance is $\left\mid L - 1 \right\mid$ and others for which the distance is $\left\mid L - \left(- 1\right) \right\mid = \left\mid L + 1 \right\mid$. These distances cannot both go to $0$.

Using the definition

Recall the definition:

${\lim}_{x \rightarrow a} f \left(x\right) = L$ if and only if

For every $\epsilon > 0$, there is a $\delta > 0$ for which
for every $x$ with $0 < \left\mid x - a \right\mid < \delta$, we have $\left\mid f \left(x\right) - L \right\mid < \epsilon$.

To use the definition to show that a limit does not exist requires showing that there is an $\epsilon > 0$ for which, for every $\delta > 0$ there is an $x$ with $0 < \left\mid x - a \right\mid < \delta$, but $\left\mid f \left(x\right) - L \right\mid \ge \epsilon$.

Given $L$ (a proposed limit), choose $\epsilon = 1$.
Claim: for every $\delta > 0$ there is an $x$ with $0 < \left\mid x - 0 \right\mid < \delta$, but $\left\mid \sin \left(\frac{2 \pi}{x}\right) - L \right\mid \ge \epsilon$

Proof of claim: For every $\delta > 0$, there is an integer $n$ for which $\left\mid \frac{4}{1 + 4 n} \right\mid < \delta$ and $\left\mid \frac{4}{3 + 4 n} \right\mid < \delta$.

Let ${x}_{1} = \frac{4}{1 + 4 n}$ and ${x}_{2} = \frac{4}{3 + 4 n}$ for such an $n$.

Then $\left\mid f \left({x}_{1}\right) - L \right\mid = \left\mid 1 - L \right\mid = \left\mid L - 1 \right\mid$.

and $\left\mid f \left({x}_{2}\right) - L \right\mid = \left\mid - 1 - L \right\mid = \left\mid L + 1 \right\mid$.

Claim: we cannot have both $\left\mid L + 1 \right\mid < 1$. and $\left\mid L + 1 \right\mid < 1$.

Suppose $\left\mid L + 1 \right\mid < 1$. Then $- 1 < L + 1 < 1$.
But his implies that $- 3 < L - 1 < - 1$, so $\left\mid L - 1 \right\mid \ge 1$

This completes the proof.