1. ## Continuity

I have two homework problems that I was hoping someone could help me with:
1. Prove that f(x)= sin(1/x), where x can't equal zero and f(0)=0 is not continuous at x=0 by finding an ε for which there is no reply.

and

2. At what values of x is
f(x) = (piecewise defined) 0, if x is irrational or = sinx if x is rational

Thank you

2. For the first one see this thread

For the second one, my guess would be that it's continous at all points of the form $\displaystyle k\pi$ where $\displaystyle k \in \mathbb{Z}$ since $\displaystyle \sin(x)=0$ for these points (and only for these).

3. Originally Posted by CoraGB
I have two homework problems that I was hoping someone could help me with:
1. Prove that f(x)= sin(1/x), where x can't equal zero and f(0)=0 is not continuous at x=0 by finding an ε for which there is no reply.

and

2. At what values of x is
f(x) = (piecewise defined) 0, if x is irrational or = sinx if x is rational

Thank you
To prove $\displaystyle f(x)$ is continuous at $\displaystyle x=k\pi$ for $\displaystyle k\in\mathbb{Z}$, we need to prove that $\displaystyle \forall~\epsilon>0$, $\displaystyle \exists~\delta>0$ s.t. $\displaystyle |f(x)-f(k\pi)|=|f(x)|<\epsilon$ if $\displaystyle |x-k\pi|<\delta$. To do this, take two cases:

1) $\displaystyle x$ is irrational. This is trivial though, because $\displaystyle |f(x)|=0<\epsilon$.

2.) $\displaystyle x\in\mathbb{Q}$ and $\displaystyle |x-k\pi|<\delta$. Since $\displaystyle \sin(x)$ is continuous, $\displaystyle |f(x)|=|\sin(x)|=|\sin(x)-\sin(k\pi)|<\epsilon$ when $\displaystyle |x-k\pi|<\delta$.

So $\displaystyle f(x)$ is continuous at $\displaystyle x=k\pi$ for $\displaystyle k\in\mathbb{Z}$.

4. And to prove it's continuous nowhere else, let $\displaystyle x_0\in\mathbb{Q}$. Since $\displaystyle \forall~\delta>0$, there exists an irrational number $\displaystyle x\in(x_0-\delta,x_0+\delta)$, $\displaystyle |f(x_0)-f(x)|=|f(x_0)|=|\sin(x_0)|>0$. So choosing $\displaystyle \epsilon=\frac{\sin(x_0)}{2}$ ensures that $\displaystyle f$ is not continuous at $\displaystyle x_0$.

A similar argument can be constructed if $\displaystyle x_0$ is irrational.

5. For what it is worth, another solution to 1.

You know that $\displaystyle \sin(1/x)=1$ when $\displaystyle \frac{1}{x}=\frac{\pi}{2} + 2k\pi$ where $\displaystyle k\in\mathbb{Z}$,
and $\displaystyle \sin(1/x)=-1$ when $\displaystyle \frac{1}{x}=\frac{3\pi}{2}+2k\pi$

Lets define a sequence $\displaystyle x_n=\frac{\pi}{2} +2n\pi$ where $\displaystyle n\in\mathbb{N}$.
Also the define the sequence $\displaystyle g_n=\frac{3\pi}{2} +2n\pi$, where $\displaystyle n\in\mathbb{N}$.
It is clear that $\displaystyle \lim\limits_{n\to\infty}x_n=\lim\limits_{n\to\inft y}g_n=\lim\limits_{x\to0^+}\frac{1}{x}=+\infty$
So we will substitude the sequences for $\displaystyle \frac{1}{x}$

Observe that
$\displaystyle \lim\limits_{n\to\infty}\sin(x_n)=1$
And
$\displaystyle \lim\limits_{n\to\infty}\sin(g_n)=-1$

Now we have two sequences $\displaystyle x_n$ and $\displaystyle g_n$ with the same limit as 1/x such that $\displaystyle \sin(x_n)$ and $\displaystyle \sin(g_n)$ have two different limits.
Hence $\displaystyle \lim\limits_{x\to0^+}\sin\left(\frac{1}{x}\right)$ does not exist, so $\displaystyle \lim\limits_{x\to0}\sin\left(\frac{1}{x}\right)$ does not exist either.

Hjörtur

6. Originally Posted by hjortur
For what it is worth, another solution to 1.

You know that $\displaystyle \sin(1/x)=1$ when $\displaystyle \frac{1}{x}=k\frac{\pi}{2}$ where $\displaystyle k\in\mathbb{Z}$,
and $\displaystyle \sin(1/x)=-1$ when $\displaystyle \frac{1}{x}=k\frac{3\pi}{2}$
Actually,

$\displaystyle \sin(1/x)=1$ iff $\displaystyle \frac{1}{x}=\frac{k\pi}{2}$ and $\displaystyle k\equiv1\mod4$

$\displaystyle \sin(1/x)=-1$ iff $\displaystyle \frac{1}{x}=\frac{k\pi}{2}$ and $\displaystyle k\equiv3\mod4$

7. Of course you are correct, I wonder why I didnt notice the error?
Any way corrected version above.