barksdalemc said:
Homework Statement
Show that f(x)=x^2sin(1/x) is piecewise continuous in (0,1)
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?
chaoseverlasting said:
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).
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.
