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
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.