Prove lim(sin(1/x)) as x tends to 0 does not exist

**boromir** · June 14th 2011, 06:20 AM

assume there is a limit, L. So we have that for all ${\epsilon}>0$ there exists a ${\sigma}>0$ such that $|x-0|=|x|<{\sigma} ->|sin(1/x)-L|<{\epsilon}$

Consider when x>_0 and let ${\epsilon}$ =1/2.

Then |x|=x -> $0<x<{\sigma}-> |sin(1/x)-L|<1/2$

I need a contradiction.

Thanks

**Prove It** · June 14th 2011, 06:25 AM

Do you need to use an $\displaystyle \epsilon - \delta$ proof? Couldn't you just say that since the function oscillates, it is impossible to say that $\displaystyle \sin{X}$ approaches a particular value as $\displaystyle X \to \infty$ ?

**boromir** · June 14th 2011, 06:29 AM

no I need to use epsilon delta

**Prove It** · June 14th 2011, 06:52 AM

We need an $\displaystyle L$ such that $\displaystyle \forall \epsilon > 0, \exists \delta$ such that $\displaystyle \left|\sin{\left(\frac{1}{x}\right)} - L \right| < \epsilon$ if $\displaystyle |x - 0| = |x| < \delta$ .

On any interval $\displaystyle (-\epsilon + \bar{x}, \bar{x} + \epsilon)$ we have $\displaystyle x \in (-\delta, \delta)$ such that $\displaystyle f(x) = \pm 1$ . Consequently for $\displaystyle \epsilon < 2$ we can't have $\displaystyle \left|\sin{\left(\frac{1}{x}\right)} - L\right| < \epsilon$ for any $\displaystyle L$ and any $\displaystyle \delta > 0$ . Thus we can not satisfy the definition no matter what we choose for $\displaystyle L$ .

**boromir** · June 14th 2011, 07:10 AM

Sorry, I don't quite understand. what x bar? Do you mean L. Then I don't know why f(x) must take the values + or - 1 in any (-delta,delta). Lastly you seem to imply L must be eith + or -1 meaning the distance between sin(1/x) and L will be 2 at some point. Why is this the case?

**Prove It** · June 14th 2011, 07:14 AM

$\displaystyle \bar{x}$ is some $\displaystyle x$ close to $\displaystyle 0$ .

The proof basically states that for values infinitessimally close to $\displaystyle x = 0$ , it can be said that $\displaystyle f(x)$ is equal to both $\displaystyle 1$ and $\displaystyle -1$ , which is enough to show that you can not "zoom in" on an $\displaystyle L$ as you make $\displaystyle \epsilon$ and $\displaystyle \delta$ smaller.

**theodds** · June 14th 2011, 07:16 AM

Consider the sequence $x_n = \frac 2 {n\pi}$ , for $n = 1, 2, ...$ .

**HallsofIvy** · June 14th 2011, 07:27 AM

In fact, consider the two sequences $x_n= \frac{1}{2n\pi}$ and $x_m= \frac{1}{n\pi}$ . You need to know that f(x) goes to L as x goes to a if and only if f(xn) goes to L for every sequence {xn} that goes to L.

**boromir** · June 14th 2011, 08:09 AM

You need to know that f(x) goes to L as x goes to a if and only if f(xn) goes to L for every sequence {xn} that goes to L. Is that a general theorem?

Don't know how you know 1/2npi goes to L but I will roll with it for now

sin(2npi)=0 so need delta such that if 0<|1/(2npi)|<delta then |-L|=|L|< e. But we can make e arbitarily small so it can be made less than |L| which is a contradiction. So f((1/2npi))has no limit so f(x) has no limit. Am I correct?

**boromir** · June 15th 2011, 02:07 AM

I've done more working

**HallsofIvy** · June 15th 2011, 02:28 AM

Given any $\delta> 0$ , there exist an integer n such that $\frac{2}{n\pi}< \delta$ . That means that every number from $-\frac{2}{n\pi}$ to $+\frac{2}{n\pi}$ are less than $\delta$ . For those values of x, sin(1/x) becomes $-sin(n\pi/2)= -1$ and $sin(n\pi/2)= 1$ . That is, no matter how small $\delta$ is, there will exist values of x less than $\delta$ such that sin(1/x) takes on values of 1 and -1. You can't guarentee any $\epsilon$ less than 2.

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