Define $G(x)=\int_0^xg(x),$ where g is given by the following:

$$g(x) = \begin{cases} \sin\frac{2}{x} & \textrm{ if $x\ne 0$} \\ 0 & \textrm{ if $x =0$} \\ \end{cases} $$

Is $G$ continuous on $[0,\frac{1}{\pi}]$?

Is $G$ differentiable on $[0,\frac{1}{\pi}]$?

Now I am aware that if I prove differentiability then continuity comes for the ride.

So my question is whats the best way to prove $G$ is differentiable.

If $$G(x) = \begin{cases} -\cos\frac{2}{x} & \textrm{ if $x\ne 0$} \\ 0 & \textrm{ if $x =0$} \\ \end{cases} $$

should I use the different quotient and show that $G$ is differentiable at all $a$.

Or is there some other more clear concise manner for proving this?

I would appreciate any tips or nudges in the right direction

  • $\begingroup$ I edited your post with the back slash \ for $\sin x, \cos x$ $\endgroup$
    – DeepSea
    Apr 16 '16 at 0:26
  • $\begingroup$ But G is [b]not[/b] equal to the formula you give! The integral of "sin(f(x))" depends on f(x). You cannot simply ignore the "1/x". $\endgroup$
    – user247327
    Apr 22 '17 at 15:10

G is continuous at 0.

$-1\leq g(x) \leq 1\\ -x\leq G(x) \leq x\\ \lim_\limits{x\to 0} G(x) = 0 = G(0)$

By the squeeze theorem.

But $G(x)$ is not differentiable at 0.

$\lim_\limits{h\to 0} \dfrac{G(h) - G(0)}{h}$ does not exist.

For any delta, there exists an h less than delta such that $g(h) = 1$ and there also exists an h less than delta such that $g(h) = -1$.

  • $\begingroup$ What is your point with the last line? $\endgroup$
    – B ry
    Apr 15 '16 at 23:53
  • $\begingroup$ If you want to make an epsilon \ delta proof of the existence of the limit, you would say $|h|<\delta \implies |g(h)|<\epsilon.$ But I say, for $|h|<\delta, g(h)$ could be as large as 1 and as small as -1. $\endgroup$
    – Doug M
    Apr 15 '16 at 23:59
  • $\begingroup$ I must be missing something, from your initial response we know the G(x) is continuous on the interval but it is not differentiable at 0, thus I could conclude from that point that G(x) is continuous but not differentiable, why do I need to establish things about g(x)? $\endgroup$
    – B ry
    Apr 16 '16 at 0:08
  • $\begingroup$ g(x) is the derivative of G(x) at all points other than 0. $\endgroup$
    – Doug M
    Apr 16 '16 at 0:09
  • $\begingroup$ so the added point doesn't plug that hole in the graph? it being piecewise doesn't change anything $\endgroup$
    – B ry
    Apr 16 '16 at 0:14

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