Diffrentiating Piecewise Function

1. Oct 27, 2009

RPierre

1. The problem statement, all variables and given/known data
Let $$L\in R$$ and define

$$h(x) = \begin{cases} sin(1/x) & \text{ if } x \neq 0 \\ L & \text{ if } x = 0 \end{cases}$$

Prove that h is not continuous at 0

2. Relevant equations

My Professor gave us the hint "Prove by contradiction, taking $$\epsilon = 1/2$$"

3. The attempt at a solution

I didn't see how this hint was relevant, but it seems too easy. All I did was say since a function is continuous at a point if it is differentiable at that point, then take 0. then h(x) = L and by first principles

lim h->0 f(x +h) - f(x) / h = L/0 which does not exist, and therefore the function is not continuous.

Why would he give that hint, and the piecewise function, when it could be proven so simply? Thanks in advance.

2. Oct 27, 2009

lurflurf

You are correct that differentiability implies continuity, but the converse is false. Lack of differentiability does not imply lack of continuity, consider |x| which is continuous when x=0 dispite failing to be differentiable. For continuity consider the family of punctured disks (-n pi,n pi)\0 where n is a positive integer.

Last edited: Oct 27, 2009
3. Oct 28, 2009

HallsofIvy

Staff Emeritus
Note that sin(1/x)= 1 for $x= 1/(2n\pi)$ for any integer n and that sin(1/x)= -1 for $x= 1/((2n+1)\pi)$ for any integer n.