Let:

Determine whether or not http://webwork.asu.edu/webwork2_file...08e14e50e1.png exists.

Yes or No?

I think it is yes because the derivative of 0 is just 0 so f'(0) would still exist at 0?

Am i correct or no?

Thank you

Chad

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- September 18th 2009, 06:09 PMmybrohshi5Determine whether or not f'(0) exists.
Let:

Determine whether or not http://webwork.asu.edu/webwork2_file...08e14e50e1.png exists.

Yes or No?

I think it is yes because the derivative of 0 is just 0 so f'(0) would still exist at 0?

Am i correct or no?

Thank you

Chad - September 18th 2009, 06:20 PMJose27
- September 18th 2009, 06:29 PMmybrohshi5
thank you.

so then for

http://webwork.asu.edu/webwork2_file...39a731d6e1.png

will it also be no because as the lim as x-->0 3x^3 sin 1/x will not exists so, f'(0) would not exist correct? - September 18th 2009, 06:41 PMJose27
- September 18th 2009, 07:48 PMmybrohshi5
can i ask a quick question about these two problems?

you said sin(1/x) is bounded so therefore it does exist. why is sin(1/x) bounded and sin(5/x) not bounded which therefore does not make sin(5/x) exist?

From looking at the graphs of both they both seem to be bounded.

I guess im just kind of confused on why one exists and the other does not and i have a test coming up this week so im just trying to make sure i understand everything clearly.

Any more help or explanations would be much appreciated.

Thank you

Chad - September 18th 2009, 08:03 PMKrizalid
those are bounded.

read again post #2, it's different from post #4. - September 18th 2009, 08:23 PMmybrohshi5
ok i just looked over both of those posts.

so for xsinx^(5/x) why was the lim x-->0 f(0 + t) - f(0) / t used

and then for 3x^3sin(1/x) why was the lim x-->0 3t^3sin(1/t)/t used?

I mean like why are two different things used for similar functions? - September 19th 2009, 08:29 AMJose27