prove that the function f(x)=....has a derivative at x=0

- anonymous

prove that the function f(x)=....has a derivative at x=0

- chestercat

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- anonymous

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

- anonymous

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?

- KingGeorge

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

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- anonymous

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?

- anonymous

*derivative at 0

- KingGeorge

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

- anonymous

oh, it's actually x²cos(1/x)

- 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.

- KingGeorge

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

- anonymous

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

- 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 \[\lim_{x\to 0} x^2\cos\left(\frac{1}{x}\right)=0.\]

- 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.

- anonymous

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

- KingGeorge

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

- KingGeorge

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

- anonymous

"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?

- 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.

- 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.

- anonymous

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.

##### 1 Attachment

- anonymous

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

- anonymous

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

- anonymous

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

- ZeHanz

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

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