## Bomull 2 years ago prove that the function f(x)=....has a derivative at x=0

1. Bomull

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

2. Bomull

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

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

4. Bomull

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

*derivative at 0

6. KingGeorge

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

7. Bomull

oh, it's actually x²cos(1/x)

8. KingGeorge

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

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

10. Bomull

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

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

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

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

14. KingGeorge

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

15. KingGeorge

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

"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

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

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

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

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

21. Bomull

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

22. Bomull

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

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