Bomull Group Title prove that the function f(x)=....has a derivative at x=0 one year ago one year ago

1. Bomull Group Title

$f(x)=x²\cos \frac{ 1 }{ 2 }, when x \neq 0; 0, when x=0$

2. Bomull Group Title

so a graphing calculator shows that x²cos(1/2) fluctuates near 0, and I believe it might not be continuous at 0. So the problem is to say that the piecewise function is defined and continuous and therefore has a derivative?

3. KingGeorge Group Title

Is that function typed correctly? I'm getting a very smooth looking parabola as my graph for that.

4. Bomull Group Title

you're right, I must have made a mistake before. but still the problem would be that if f(x) wasn't defined to be 0 at 0, it wouldn't be continuous and have a derivate at 0?

5. Bomull Group Title

*derivative at 0

6. KingGeorge Group Title

Could it have been $x^2\cos\left(\frac{1}{2x}\right)$

7. Bomull Group Title

oh, it's actually x²cos(1/x)

8. KingGeorge Group Title

Ah. Well, to show that it's differentiable, you basically have to show that it's continuous, and that the limit $\lim_{h\to0} \frac{f(0+h)-f(0)}{h}$exists.

9. KingGeorge Group Title

To show continuity, you need to show$\lim_{x\to 0} x^2\cos\left(\frac{1}{x}\right)=0.$

10. Bomull Group Title

so cos (1/x) at x->0 would be undefined because of (1/0), right? but then I don't know how to go further

11. KingGeorge Group Title

What you could do, is since $$\cos(x)$$ is bounded by -1 and 1, when you multiply by $$x^2$$, that term will dominate. So $\lim_{x\to 0} x^2\cos\left(\frac{1}{x}\right)=0.$

12. KingGeorge Group Title

In the first limit, $\lim_{h\to0} \frac{f(0+h)-f(0)}{h} =\lim_{h\to0} \frac{h^2\cos\left(\frac{1}{h}\right)}{h}$We just showed that the numerator goes to 0, so we can use L'hopital's on this.

13. Bomull Group Title

L'hopital's rule comes only later in the course!

14. KingGeorge Group Title

In that, case, try multiplying by $\Large\frac{\frac{1}{h}}{\frac{1}{h}}$

15. KingGeorge Group Title

Then we get $\lim_{h\to0}\frac{h\cos\left(\frac{1}{h}\right)}{1}=\lim_{h\to0}h\cos\left(\frac{1}{h}\right)$

16. Bomull Group Title

"What you could do, is since cos(x) is bounded by -1 and 1, when you multiply by x2, that term will dominate." but does it not matter that 1/0 would be undefined?

17. KingGeorge Group Title

But you aren't dividing by 0. You're just getting very close. So you have $\cos(\text{A very large number})$ which is bounded by -1 and 1.

18. KingGeorge Group Title

I've got to go now. Hopefully that made enough sense. To do that very last limit, use the same fact that cosine is bounded.

19. robtobey Group Title

Refer to the attached plot of$\left\{x^2 \cos \left(\frac{1}{x}\right),\sin \left(\frac{1}{x}\right)+2 x \cos \left(\frac{1}{x}\right)\right\}$in blue and red respectively.

20. robtobey Group Title

Sorry, the tiff graphic above is not viewable unless your have access to Apple's Safari browser or perhaps photoshop.

21. Bomull Group Title

@robtobey I can see it on Ubuntu no problem. Now I just need to figure out what youre trying to say ;)

22. Bomull Group Title

why is $\lim_{h \rightarrow 0}\frac{ f(0+h)-f(0) }{ h }=\lim_{h \rightarrow 0}\frac{ h²\cos(\frac{ 1 }{ h }) }{ h }$ ?

23. ZeHanz Group Title

Because $$f(0+h)=f(h)=h^2cos(\frac{1}{h})$$ (just replace x with h) Also, $$f(0)=0$$, so$\lim_{h \rightarrow 0}\frac{ f(0+h)-f(0) }{ h }=\lim_{h \rightarrow 0}\frac{ f(h)-f(0) }{ h }=\lim_{h \rightarrow 0}\frac{ h^2\cos(\frac{1}{h})-0 }{ h }=\lim_{h \rightarrow 0}\frac{ h^2\cos(\frac{1}{h})}{ h }$