# Determine whether or not f'(0) exists.

• Sep 18th 2009, 07:09 PM
mybrohshi5
Determine 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

• Sep 18th 2009, 07:20 PM
Jose27
Quote:

Originally Posted by mybrohshi5
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

Remember that the derivative is defined in terms of a limit, so you have:

$f'(0)= \lim_{t \rightarrow 0} \frac{f(0+t) - f(0)}{t} = \lim_{t \rightarrow 0} \frac{f(t)}{t}= \lim_{t \rightarrow 0} [\sin (\frac{5}{t} )]$ which doesn't exist, therefore $f'(0)$ doesn't exist.
• Sep 18th 2009, 07:29 PM
mybrohshi5
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?
• Sep 18th 2009, 07:41 PM
Jose27
Quote:

Originally Posted by mybrohshi5
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?

Be careful:

$f'(0)= \lim_{t \rightarrow 0} \frac{3t^3 \sin( \frac{1}{t} ) }{t} =\lim_{t \rightarrow 0} 3t^2 \sin (\frac{1}{t} )$ and since $\sin (\frac{1}{t} )$ is bounded, the limit is $0$ so $f'(0)=0$
• Sep 18th 2009, 08:48 PM
mybrohshi5

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

• Sep 18th 2009, 09:03 PM
Krizalid
those are bounded.

read again post #2, it's different from post #4.
• Sep 18th 2009, 09:23 PM
mybrohshi5
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?
• Sep 19th 2009, 09:29 AM
Jose27
Quote:

Originally Posted by mybrohshi5
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?

$\lim_{t \rightarrow 0} \frac{f(0+t)-f(0)}{t}= \lim_{t \rightarrow 0} \frac{3t^3 \sin ( \frac{1}{t} )}{t}$. It's the same. The first limit doesn't exist because although $\sin( \frac{1}{t} )$ is bounded the limit doesn't exist, but for the second limit, the term $3t^2$ dominates and it goes to $0$