Showing continutiy

1. Dec 14, 2006

barksdalemc

1. The problem statement, all variables and given/known data
Show that f(x)=x^2sin(1/x) is piecewise continuous in (0,1)

3. The attempt at a solution

I'm trying to show continuity in (0,1) so I need to show limit as function approaches any given Xo is the value of the function itself, ie. |X^2sin(1/X) - Xo^2sin(1/Xo)|< eps. Can anyone help from here?

2. Dec 14, 2006

jing

3. Dec 14, 2006

barksdalemc

Correct, but I am given only the interval (0,1). At the least I need to show continuous in that interval. Not sure what choice of eps for this function.

4. Dec 14, 2006

HallsofIvy

Staff Emeritus
You don't need piecewise here: the function f(x)= x2sin(1/x) not defined "piecewise" on (0, 1).

The problem is that we don't know what you are "allowed" to use. Obviously this is the product of two functions both of which are continuous on (0,1). Are you saying that you are required to use the definition of limit?
It might help to remember that $-1\le sin(\theta )\le 1$ for all $\theta$.

5. Dec 14, 2006

StatusX

Just use/prove the fact that the composition and product of continuous functions is continuous, and that sin(x), x^2, and 1/x are continuous on (0,1). And please be cleaner with notation- I'm assuming you mean x2sin(1/x) and not x2sin(1/x)

Last edited: Dec 14, 2006
6. Dec 14, 2006

barksdalemc

Yes I am required to use the definition of the limit. I was going to use the fact that sin x is bounded. Can I say |X^2sin(1/X) - Xo^2sin(1/Xo)| is always less than or equal to |X^2-Xo^2|? Then I can choose an epsilon that works for f(x)=X^2 to show continuity.

7. Dec 14, 2006

chaoseverlasting

The function is $$x^2 sin(\frac{1}{x})$$ the derivative of which is
$$2x sin(\frac{1}{x}) -cos(\frac{1}{x})$$ which exists for 0<x<1, therefore, the function is continuous for x belonging to (0,1).

8. Dec 14, 2006

jing

Just to clarify by the interval (0,1) I presume you mean for

0< x < 1 and not 0<= x <= 1

Is 1/x continuous at x=0?

9. Dec 14, 2006

chaoseverlasting

Yes, I meant 0<x<1 as the function does not exist at x=0. I dont think 1/x is continuous cause then f(x) is infinity, and the graph of the curve f(x)=1/x at x=0 definately breaks.

10. Dec 14, 2006

maverick280857

$x^{2}$ is continuous everywhere in (0,1).

$\sin(1/x)$ is continuous at all points except when $1/x = \n \pi$ where n is a postive integer. That is it is singular at $x = 1/n\pi$ in (0,1). How many values of n satisfy this equation?

The derivative cannot be computed for all points in (0,1) for the reason given above. In fact this is conceptually wrong. You cannot compute the derivative first and judge the continuity of the original function! As an example, consider the function $|x|$ in (-1,1) (or any interval containing 0). The function is continuous everywhere in (-1,1) and for that matter in every interval of R, but its derivative is a piecewise continuous function in an interval containing zero.

Hope that helps.

barksdalemc: Don't apply the definition of limit (i.e. the so called epsilon delta criterion) mechanically. Judge the behavior of the functions involved near the point in question...quite often, its easier.

I don't know whether you're familiar with sequences (omit this if you don't). There is a neater way to do the problem formally if you realize that the continuity (or discontinuity) of the function in (0,1) hinges only the continuity of $\sin(1/x)$ in (0,1). If $f(x)$ is continuous then ${x_{n}} \rightarrow x_{0}$ implies $f({x_{n}}) \rightarrow f(x_{0})$. This in fact is a double-sided implication, so you can use it to prove or disprove the continuity of f(x). The proof of this "theorem" (some call it the sequential criterion for continuity) is slightly involved however. In our problem, you can consider a sequence ${x_{n}} = 1/n\pi$ and observe that as $n \rightarrow \infty$, $x_{n} \rightarrow 0$. Clearly, $f(x_{n}) = 0$ but $Lim_{x \rightarrow \infty} f(x)$ is undefined.

Last edited: Dec 14, 2006