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Bomull
 one year ago
Best ResponseYou've already chosen the best response.1\[f(x)=x²\cos \frac{ 1 }{ 2 }, when x \neq 0; 0, when x=0\]

Bomull
 one year ago
Best ResponseYou've already chosen the best response.1so 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
 one year ago
Best ResponseYou've already chosen the best response.3Is that function typed correctly? I'm getting a very smooth looking parabola as my graph for that.

Bomull
 one year ago
Best ResponseYou've already chosen the best response.1you'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?

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.3Could it have been \[x^2\cos\left(\frac{1}{2x}\right)\]

Bomull
 one year ago
Best ResponseYou've already chosen the best response.1oh, it's actually x²cos(1/x)

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.3Ah. 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
 one year ago
Best ResponseYou've already chosen the best response.3To show continuity, you need to show\[\lim_{x\to 0} x^2\cos\left(\frac{1}{x}\right)=0.\]

Bomull
 one year ago
Best ResponseYou've already chosen the best response.1so 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
 one year ago
Best ResponseYou've already chosen the best response.3What 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
 one year ago
Best ResponseYou've already chosen the best response.3In 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.

Bomull
 one year ago
Best ResponseYou've already chosen the best response.1L'hopital's rule comes only later in the course!

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.3In that, case, try multiplying by \[\Large\frac{\frac{1}{h}}{\frac{1}{h}}\]

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.3Then we get \[\lim_{h\to0}\frac{h\cos\left(\frac{1}{h}\right)}{1}=\lim_{h\to0}h\cos\left(\frac{1}{h}\right)\]

Bomull
 one year ago
Best ResponseYou've already chosen the best response.1"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
 one year ago
Best ResponseYou've already chosen the best response.3But 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
 one year ago
Best ResponseYou've already chosen the best response.3I've got to go now. Hopefully that made enough sense. To do that very last limit, use the same fact that cosine is bounded.

robtobey
 one year ago
Best ResponseYou've already chosen the best response.0Refer 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.

robtobey
 one year ago
Best ResponseYou've already chosen the best response.0Sorry, the tiff graphic above is not viewable unless your have access to Apple's Safari browser or perhaps photoshop.

Bomull
 one year ago
Best ResponseYou've already chosen the best response.1@robtobey I can see it on Ubuntu no problem. Now I just need to figure out what youre trying to say ;)

Bomull
 one year ago
Best ResponseYou've already chosen the best response.1why is \[\lim_{h \rightarrow 0}\frac{ f(0+h)f(0) }{ h }=\lim_{h \rightarrow 0}\frac{ h²\cos(\frac{ 1 }{ h }) }{ h }\] ?

ZeHanz
 one year ago
Best ResponseYou've already chosen the best response.0Because \(f(0+h)=f(h)=h^2cos(\frac{1}{h})\) (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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