1. ## Correctly shown?

Just checking this now :\ I had to show that f is not differentiable at 0 so:

Potential Proof:

$\displaystyle |\frac{f(0+h) - f(0)}{h} - 0 | = |\frac{f(0+h) - f(0)}{h} |$

$\displaystyle =\fraq{{|f(h)|}} {{|h|}}$

$\displaystyle \le$ $\displaystyle \fraq{{h}} {{|h|}}$

$\displaystyle = 1$

I think I have made a big booboo and gone completely wrong, here is the whole of the question (its not an exam question, but a simple question from a group that our lecturer has given us to try and do over summer before anyone asks)

PS: It is spose to show f(h) over the modulus of h and in the line below, h over modulus of h

2. Originally Posted by ramdrop
Just checking this now :\ I had to show that f is not differentiable at 0 so:

Potential Proof:

$\displaystyle |\frac{f(0+h) - f(0)}{h} - 0 | = |\frac{f(0+h) - f(0)}{h} |$

$\displaystyle =\fraq{{|f(h)|}} {{|h|}}$

$\displaystyle \le$ $\displaystyle \fraq{{h}} {{|h|}}$

$\displaystyle = 1$

I think I have made a big booboo and gone completely wrong, here is the whole of the question (its not an exam question, but a simple question from a group that our lecturer has given us to try and do over summer before anyone asks)

PS: It is spose to show f(h) over the modulus of h and in the line below, h over modulus of h
Similar to this problem on another forum.

Show f(x) = { x/2 if x rational , x if x irrational is not differentiable at 0

Basic idea is that there are two possible (secant) slopes: 0 and 1, and you can always find values of x satisfying both in any (-delta, delta) \ {0}, delta > 0.

I don't entirely follow your potential proof. Shouldn't you be subtracting L instead of 0 inside the first absolute value?

3. Ah looked through my proof, it is very wrong so, I looked at the other forum and got a few ideas and tried it that way and reached 2 inequalities that contradicted each other, now starting with part A (seeing as I did b first, d'oh)

I would try:

Let x'n = sequence of irrational numbers tending to x

f(xn) = 0 -> 0 $\displaystyle \ne$ x

But that just goes wrong, maybe I have to set it up differently, maybe hmm, well i dunno guess I must look harder

4. Originally Posted by ramdrop
Just checking this now :\ I had to show that f is not differentiable at 0 so:

Potential Proof:

$\displaystyle |\frac{f(0+h) - f(0)}{h} - 0 | = |\frac{f(0+h) - f(0)}{h} |$

$\displaystyle =\fraq{{|f(h)|}} {{|h|}}$

$\displaystyle \le$ $\displaystyle \fraq{{h}} {{|h|}}$

$\displaystyle = 1$

I think I have made a big booboo and gone completely wrong, here is the whole of the question (its not an exam question, but a simple question from a group that our lecturer has given us to try and do over summer before anyone asks)

PS: It is spose to show f(h) over the modulus of h and in the line below, h over modulus of h
Proof.
a) Let $\displaystyle \{x_n\}$ be a sequence such that $\displaystyle x_n\rightarrow 0 (n\rightarrow \infty)$. To show $\displaystyle f(x_n)\rightarrow f(0) (n\rightarrow \infty).$

$\displaystyle |f(x_n)-f(0)|=|f(x_n)|\leq |x_n|\rightarrow 0 (n\rightarrow \infty)$

b)Aim to show $\displaystyle \lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x-0}$ doesn't exist, i.e., $\displaystyle \lim_{x\rightarrow 0}\frac{f(x)}{x}$ doesn't exist. From the definition of LIMIT, let $\displaystyle a$ be a real number. To show

$\displaystyle \exists \varepsilon_0>0, \forall \delta>0, \exists |x|<\delta$ such that $\displaystyle |\frac{f(x)}{x}-a| \geq \varepsilon_0$.

Actually, it is easy to find such $\displaystyle \varepsilon_0$:just set it to be $\displaystyle \frac{a}{2}$, noting that $\displaystyle x$ should be irrational.

5. Thanks man, is that like the full proof seems really short