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

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

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?

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

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?

*derivative at 0

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

oh, it's actually x²cos(1/x)

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.

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

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

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

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.

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

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

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

"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?

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.

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.

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.

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

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

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

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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