Show that g′ is discontinuous at x = 0 and that g′ takes on values close to ±1 arbitrarily near 0. When g(x) = $x^2 sin(1/x)$
Show that g′ is discontinuous at x = 0 and that g′ takes on values close to ±1 arbitrarily near 0.
-
1What's the point of us doing your homework? – 2012-11-08
1 Answers
If you define $f(x)$ to be $0$ at $x=0$, and $x^2\sin(1/x)$ when $x\ne 0$, it turns out that $f'(0)$ exists and $f'(0)=0$. That is not hard to show from the definition of the derivative. We will not do it, partly because it is not relevant to the problem.
The wording of your problem is somewhat unusual. It is easier to reverse it, by (i) showing that $g'(x)$ takes on values arbitrarily close to $\pm 1$ arbitrarily close to $x=0$ and (ii) deducing that $g'(x)$ is not continuous at $0$.
We will do mostly the part you undoubtedly know how to do. By the Product Rule and the Chain Rule, we have, for $x\ne 0$, that $g'(x)=2x\sin(1/x)+(x^2)(-1/x^2)\cos(1/x)=2x\sin(1/x)-\cos(1/x).$ Observe that $|\sin(1/x)|\le 1$. Can you conclude that $2x\sin(1/x)$ has small absolute value near $0$? Now look at the second term $-\cos(1/x)$. Can you find $x$ very very near $0$ at which $\cos(1/x)=1$? What about $-1$?