Let:
Determine whether or not 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
Remember that the derivative is defined in terms of a limit, so you have:
$\displaystyle f'(0)= \lim_{t \rightarrow 0} \frac{f(0+t) - f(0)}{t} = \lim_{t \rightarrow 0} \frac{f(t)}{t}= \lim_{t \rightarrow 0} [\sin (\frac{5}{t} )]$ which doesn't exist, therefore $\displaystyle f'(0)$ doesn't exist.
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
$\displaystyle \lim_{t \rightarrow 0} \frac{f(0+t)-f(0)}{t}= \lim_{t \rightarrow 0} \frac{3t^3 \sin ( \frac{1}{t} )}{t}$. It's the same. The first limit doesn't exist because although $\displaystyle \sin( \frac{1}{t} )$ is bounded the limit doesn't exist, but for the second limit, the term $\displaystyle 3t^2$ dominates and it goes to $\displaystyle 0$