# Thread: Show that the function is discontinuous for x = 0

1. ## Show that the function is discontinuous for x = 0

Let:

$f(x)=\left\{\begin{matrix} \mbox{sin}(\pi/x) & \mbox{} x \neq 0 \\ 0& \mbox{} x=0\end{matrix}\right$

Show that f is not continuous for x = 0

For both $\lim_{x \to 0^+} \ f(x)$ and $\lim_{x \to 0^-} \ f(x)$ we have that they are a number [-1,1]. We also see that $f(0)$ is not defined.

How to proceed?
What I know is that if $\lim_{x \to c^+} \ f(x) = \lim_{x \to c^-} \ f(x) = f(c)$, then the function is continuous. But what role does the $0, x=0$ have?

2. First, did you graph the function near zero?
Had you, you would have seen that the graph osculates wildly between $1~\&~-1$ near zero.
This is such a famous graph it is known as the topologist sine curve.
So no limit exist at $x=0$.

$x_n=\frac{1}{2n}$
$x_m=\frac{1}{2m+1}$

4. Originally Posted by Plato
First, did you graph the function near zero?
I did, but I think I have to show this by other means than just by a graph.

5. Look at the next reply.

6. Originally Posted by zzzoak
$x_n=\frac{1}{2n}$
$x_m=\frac{1}{2m+1}$
Someone did actually try to explain a method to me, by using these sets, to solve the problem. However, I understood nothing of it... Would you care to explain further, and maybe this time I'll understand more?

7. Originally Posted by jenkki
Someone did actually try to explain a method to me, by using these sets, to solve the problem. However, I understood nothing of it... Would you care to explain further, and maybe this time I'll understand more?
Actually the sequences should be like this.
$(x_n)=\dfrac{2}{4n-3}~\&~(y_n)=\dfrac{2}{4n-1}$

Notice that $\left( {\forall n} \right)\left[ {f\left( {x_n } \right) = 1\;\& \;f\left( {y_n } \right) = - 1} \right]$.

Also, ${\left( {x_n } \right) \to 0\;\& \;\left( {y_n } \right) \to 0}$.

That means that $f$ cannot have a limit at $x=0$.

8. I'm following you until the last sentence:

Originally Posted by Plato
That means that $f$ cannot have a limit at $x=0$.

Why is this so?
Does it have something to do with the fact than when ${\left( {x_n } \right) \to 0$, then it will be like taking $sin(\pi/0)$ ? A mathematical explanation that this means that $f(x)$ is discontinuous for x = 0 would be great

9. Plato is referring to the sequential definition of continuity - Continuous function - Wikipedia, the free encyclopedia

10. Originally Posted by jenkki
Let:

$f(x)=\left\{\begin{matrix} \mbox{sin}(\pi/x) & \mbox{} x \neq 0 \\ 0& \mbox{} x=0\end{matrix}\right$

Show that f is not continuous for x = 0

For both $\lim_{x \to 0^+} \ f(x)$ and $\lim_{x \to 0^-} \ f(x)$ we have that they are a number [-1,1]. We also see that $f(0)$ is not defined.
Quite the contrary! f(0) is defined- it is 0.

How to proceed?
What I know is that if $\lim_{x \to c^+} \ f(x) = \lim_{x \to c^-} \ f(x) = f(c)$, then the function is continuous. But what role does the $0, x=0$ have?
If it were true that f(0) were undefined, then you would be done. the limit cannot equal f(0) if f(0) is undefined! But it is defined so in order not to be continuous, either the limit does not exist or the limit exists but is not equal to 0.

What is $\lim_{x\to 0} sin(\pi/x)$? You asset that the two one sides limits "are a number in [-1, 1]" but you don't show that. What are those limits? In order that the limit exist the two one-sided limits must be the same.