I am stuck at this one question, mainly because I can't understand the question. Any help will be appreciated!!

Qn

In the following question, c=3

http://img28.imageshack.us/img28/6776/cma2q2.gif

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- Apr 12th 2010, 06:51 AMtottijohnQuestion on Differentiation
I am stuck at this one question, mainly because I can't understand the question. Any help will be appreciated!!

__Qn__

In the following question, c=3

http://img28.imageshack.us/img28/6776/cma2q2.gif - Apr 12th 2010, 07:58 AMHallsofIvy
Because it is talking about the derivative, this $\displaystyle Q(h)$ is the "difference quotient", $\displaystyle \frac{f(0+h)- f(0)}{h}$.

Because f, here, is defined by different formulas for x< 0 and x> 0, you must use the limits "from the right" and "from the left".

As h approaches 0 "from the left, 0+ h= h< 0 so $\displaystyle f(0+h)= f(h)= h+ h^2$. f(0)= sin(3(0))= 0 so, for h< 0, $\displaystyle \frac{f(0+h)- f(0)}{h}= \frac{h+ h^2}{h}$.

$\displaystyle \lim_{h\to 0^-} Q(h)= \lim_{h\to 0}\frac{h+ h^2}{h}$.

As h approaches 0 "from the right", 0+ h= h> 0 so $\displaystyle f(0+h)= f(h)= sin(3h)$. for h> 0, $\displaystyle \frac{f(0+h)- f(0)}{h}= \frac{sin(3h)}{h}$.

$\displaystyle \lim_{h\to 0^+}Q(h)= \lim_{h\to 0}\frac{sin(3h)}{h}$.

To find that limit, you will need to remember a limit formula for $\displaystyle \lim_{x\to 0}\frac{sin(x)}{x}$ and use the fact that $\displaystyle \frac{sin(3h)}{h}= 3\frac{sin(3h)}{(3h)}$.

The function, f, will be differentiable at x= 0 if and only if those two one sided limits are the same. - Apr 13th 2010, 01:58 AMtottijohn
Thanks for your help!

Wow, the maths syntax is hard to use.

Anyway, the answer is 3 since lim h->0(sinh/h)=1