Correctly shown?

**ramdrop** · July 23rd 2010, 04:58 AM

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

Potential Proof:

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

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

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

$= 1$

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

**undefined** · July 23rd 2010, 05:08 AM

Originally Posted by ramdrop

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

Potential Proof:

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

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

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

$= 1$

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?

**ramdrop** · July 23rd 2010, 05:27 AM

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 $\ne$ x

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

**mchaelzang** · July 23rd 2010, 07:25 AM

Originally Posted by ramdrop

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

Potential Proof:

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

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

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

$= 1$

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 $\{x_n\}$ be a sequence such that $x_n\rightarrow 0 (n\rightarrow \infty)$ . To show $f(x_n)\rightarrow f(0) (n\rightarrow \infty).$

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

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

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

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

**ramdrop** · July 23rd 2010, 12:33 PM

Thanks man, is that like the full proof

seems really short

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