MHB Piecewise Continuous and piecewise smooth functions

comfortablynumb
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I do not know to start. Here is the problem.Determine if the given function is piecewise continuous, piecewise smooth, or neither. Here $x\neq0$ is in the interval $[-1,1]$ and $f(0)=0$ in all cases.

1. $f(x)=sin(\frac{1}{x})$
2. $f(x)=xsin(\frac{1}{x})$
3. $f(x)={x}^{2}sin(\frac{1}{x})$
4. $f(x)={x}^{3}sin(\frac{1}{x})$ .
 
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comfortablynumb said:
I do not know to start. Here is the problem.Determine if the given function is piecewise continuous, piecewise smooth, or neither. Here $x\neq0$ is in the interval $[-1,1]$ and $f(0)=0$ in all cases.

1. $f(x)=sin(\frac{1}{x})$
2. $f(x)=xsin(\frac{1}{x})$
3. $f(x)={x}^{2}sin(\frac{1}{x})$
4. $f(x)={x}^{3}sin(\frac{1}{x})$ .
Surely you can do the piecewise continuous part? It's just matching up if the curves are continous. So, for example, sin(1/x) has to have two limits: [math]\lim_{x \to 0^+} f(x) = 0[/math] (since f(0) = 0) and [math]\lim_{x \to 0^-} f(x) = 0[/math].

Piecewise smooth would be if the first derivatives are continuous. Do you need help with that part?

-Dan
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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