Proving a derivative doesn't exist

[Joe Mercurio]

Ok, I give up

F(x) =
\(\displaystyle \left\{\begin{matrix} x\ast sin\frac{1}{x} & if x \neq 0\\
0 & if x = 0
\end{matrix}\right.\)

Part B of the question asks that I prove that f(x) is continuous at X = 0. No problem
1) 0 is in the domain of f(x)? Check
2) We can easily prove that the function is continuous via the squeeze theorm using y=x and y=-x to squeeze it in the middle. The limit as x approaches 0 from the left = 0 and the limit as x approaches 0 from the right = 0 which equals f(0) = 0.

Part D asks that I prove that f'(0) does not exist. My initial thoughts were that the derivative at x=0 should be zero as well due to 1) the function being continuous at all points and 2) the limit from each side approaching zero as well so the derivative of 0 = 0. But that's obviously not the correct answer.

My thoughts now are that when I do the long way of taking a derivative or: \(\displaystyle \frac{f(x+\Delta x) - f(x)}{\Delta x}\) that the equation doesn't reduce the same way it would as if 0 wasn't involved because f(x) = 0, not x*sin(1/x). So the answer for why the derivative doesn't exist would be that f(x) and f(x+\(\displaystyle \Delta x\)) are different equations altogether and that I wouldn't be able to cancel out the \(\displaystyle \Delta x\) on the bottom.

What's not helping me is that I'm unable to take the derivative (the long way) when 0 is not involved. Beyond some simple algebraic rearranging, I haven't been able to reduce this equation into what the true derivative is: sin(1/x) - (cos(1/x)/x). Mostly because I don't know how to get the (x+\(\displaystyle \Delta x\) out of the bottom like you might be able to do with the addition property.

\(\displaystyle \frac{(x+\Delta x)*\sin \frac{1}{x+\Delta x}-x*\sin \frac{1}{x}}{\Delta x}\)

 

[Bob Brown MSEE]

F'(x)



Even if you cannot calculate this, you could guess that F'(x) oscillates with ever increasing amplitude. Think about how vertical the transitions of sin(1/x) are becoming as x->0.

 

[Joe Mercurio]

Bob,

Hmm, I don't know if that addressed my question. I calculated the derivative of x*sin(1/x) as shown in your post, but that's the derivative of f(x) at all points other than x=0. I don't see how that helps me prove that the derivative of f(x) at x=0 does not exist. The function at x=0 is 0, not x*sin(1/x)

 

[Bob Brown MSEE]

The Amplitude of -cos(1/x)/x is growing without bound.
You can not squeeze it to zero. (as x->0)

 

[Bob Brown MSEE]

The function at x=0 is 0, not x*sin(1/x)

Note: Let h(x) = |x|

The value h(0)=0 but h'(0) is undefined.
Just because a function is zero at the origin, doesn't assure that the derivative is defined there.

 

[Bob Brown MSEE]

Doesn't mean f'(0) isn't 0 either!!!

Think about:

x²sin(1/x) or x³sin(1/x)

Punchline:

Lim_x->0 x³sin(1/x) = 0
Lim_x->0 (d/dx)(x³sin(1/x)) = 0 too!!!!!!!!!!!!!

So you DO need more than my argument that "The transitions of sin(1/x) are becoming increasingly vertical".
You can use the expression for f'(x) and argument about divergence of peak values.
For any local maximum that you name, I can find you one closer to the origin.
For any local maximum that you name, I can find you one closer to the origin AND bigger.

What about a delta-epsilon arguement?

. I calculated the derivative of x*sin(1/x) as shown in your post, but that's the derivative of f(x) at all points other than x=0. I don't see how that helps me

Use it for calculating δ = f'(ε)

 

[Joe Mercurio]

What about a delta-epsilon arguement?

The delta-epsilon argument just formalizes the concept of a limit but I don't think it helps me here. I think this is the solution, please tell me if it is legit or not

I want to start at the point x=0 and go a miniscule amount to the right of it, call it \(\displaystyle \Delta x\). So at
x = 0, f(x) = 0
at (x+\(\displaystyle \Delta x\)), the value of f(x+\(\displaystyle \Delta x\)) is simply \(\displaystyle \Delta x\)*sin(1/\(\displaystyle \Delta x\))

so lim as delta(x)-->0 \(\displaystyle \frac{f(x+\Delta x)-f(x)}{\Delta x}\)

plugging in: lim as delta(x)-->0 \(\displaystyle \frac{(\Delta x*\sin (\frac{1}{(x+\Delta x)})-0)}{\Delta x}\)

reducing: lim as delta(x)-->0 \(\displaystyle \sin \frac{1}{(x+\Delta x)}\)

since x = 0: lim as delta(x)-->0 \(\displaystyle \sin \frac{1}{(\Delta x)}\)

And as you take the limit as x--> 0, this function is clearly not defined.

 

[girodd]

The delta-epsilon argument just formalizes the concept of a limit but I don't think it helps me here. I think this is the solution, please tell me if it is legit or not

I want to start at the point x=0 and go a miniscule amount to the right of it, call it \(\displaystyle \Delta x\). So at
x = 0, f(x) = 0
at (x+\(\displaystyle \Delta x\)), the value of f(x+\(\displaystyle \Delta x\)) is simply \(\displaystyle \Delta x\)*sin(1/\(\displaystyle \Delta x\))

so lim as delta(x)-->0 \(\displaystyle \frac{f(x+\Delta x)-f(x)}{\Delta x}\)

plugging in: lim as delta(x)-->0 \(\displaystyle \frac{(\Delta x*\sin (\frac{1}{(x+\Delta x)})-0)}{\Delta x}\)

reducing: lim as delta(x)-->0 \(\displaystyle \sin \frac{1}{(x+\Delta x)}\)

since x = 0: lim as delta(x)-->0 \(\displaystyle \sin \frac{1}{(\Delta x)}\)



And as you take the limit as x--> 0, this function is clearly not defined.

This is the right way to do the problem. However, rather than simply say "clearly there is no limit" you should state that sin(1/deltax) takes on both of the the values 1 and -1 for deltax in any interval about 0, and hence there is no fixed L for which sin(1/deltax) is within, say, 1/4 of L for every deltax within, say, .1 of zero. This shows that the delta-epsilon definition of limit cannot be satisfied..