# Continuity

• October 21st 2009, 05:55 PM
CoraGB
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
• October 21st 2009, 06:15 PM
Jose27
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 $k\pi$ where $k \in \mathbb{Z}$ since $\sin(x)=0$ for these points (and only for these).
• October 21st 2009, 08:36 PM
redsoxfan325
Quote:

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

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

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

So $f(x)$ is continuous at $x=k\pi$ for $k\in\mathbb{Z}$.
• October 21st 2009, 08:55 PM
redsoxfan325
And to prove it's continuous nowhere else, let $x_0\in\mathbb{Q}$. Since $\forall~\delta>0$, there exists an irrational number $x\in(x_0-\delta,x_0+\delta)$, $|f(x_0)-f(x)|=|f(x_0)|=|\sin(x_0)|>0$. So choosing $\epsilon=\frac{\sin(x_0)}{2}$ ensures that $f$ is not continuous at $x_0$.

A similar argument can be constructed if $x_0$ is irrational.
• October 22nd 2009, 12:09 PM
hjortur
For what it is worth, another solution to 1.

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

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

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

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

Hjörtur
• October 22nd 2009, 01:10 PM
redsoxfan325
Quote:

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

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

Actually,

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

$\sin(1/x)=-1$ iff $\frac{1}{x}=\frac{k\pi}{2}$ and $k\equiv3\mod4$
• October 22nd 2009, 02:18 PM
hjortur
Of course you are correct, I wonder why I didnt notice the error?
Any way corrected version above.