Derivatives of functions composition: $\lim_{x\rightarrow 8}\frac{\root{3}\of{x} - 2}{\root{3}\of{3x+3}-3}$ I have to calculate the folowing:
$$\lim_{x\rightarrow 8}\frac{\root{3}\of{x} - 2}{\root{3}\of{3x+3}-3}$$
I am not allowed to used anything else than the definition of the derivative of a function $f(a)$ with respect to $x$, that is 
$$f'(a) = \lim_{x\rightarrow a} \frac{f(x) - f(a)}{x - a}$$
After some thinking I found out we could define two function $g(x) = \frac{x-3}{3}$ and $f(x) = \root{3}\of{3x+3}$ edit: this should actually be $h(x)$
Then $h(g(x)) = \root{3}\of{x}$, and $h(8) = 3\\$ and finally: $h(g(8)) = 2$
If $f(x)$ is the first function (at the beginning), I could now rewrite it's limit as:
$$
\lim_{x\rightarrow 8} \frac{h(g(x)) - h(g(8))}{h(x) - h(8)}
$$ (1)

My question is:
If we note $c(x) = (h(g(x)))$, Does it make sense to write $h(x) \rightarrow h(8)$ under the $lim$ symbol instead of $x\rightarrow 8$? Is the above limit equal to the derivative of $c$ with respect to $h$? Does this even make sense since $h$ is applied "after" $g$?

If I just apply the chain rule:
$$\frac{dc}{dx} = \frac{dc}{dh}\frac{dh}{dx} \Leftrightarrow \frac{dc}{dh} = \frac{dc}{dx}\frac{dx}{dh}$$
then I find a wrong limit:
We are looking for $\frac{dc}{dh}$. So we can calculate 
$$
   \frac{dc}{dx} = \frac{1}{3\root{3}\of{(3x+3)^2}} \\
   \frac{dh}{dx} = \frac{3}{3\root{3}\of{(3x+3)^2}} \\
   \Rightarrow \frac{dc}{dh} = \frac{1}{3}
$$
So the limit should be one third, but it actually turns out to be $\frac{3}{4}$.
Why can't we write the limit $(1)$ as $\frac{dc}{dh}$, and more importantly, how can I get the correct limit using the definition of the derivative?

PS: I'm not sure if it's even possible because the exercise doesn't say which method to use (I know it's possible using the factorization of $a^3 - b^3$). I just find it would be really cool to solve the exercise this way

Edit
I have made a mistake on the derivative of $h(g(x)) = \root{3}\of{x}$. I used the chain rule $c'(x) = h'(g(x))\cdot g'(x)$ and that led to a mistake. I should simply have derived $c(x) = \root{3}\of{x}$. So the actual derivative is $\frac{1}{3\root{3}\of{x^2}}$. This leads to the correct limit, as showed in farruhota's answer.
 A: We don't need composite function, indeed we can use that
$$\lim_{x\rightarrow 8}\frac{\root{3}\of{x} - 2}{\root{3}\of{3x+3}-3}
=\lim_{x\rightarrow 8}\frac{\root{3}\of{x} - 2}{x-8}\frac{x-8}{\root{3}\of{3x+3}-3}$$
and use derivative definition for $f(x)=\root{3}\of{x}$ and $g(x)=\root{3}\of{3x+3}$.
A: Assuming that the derivative of the numerator and denominator exist and are finite, then we can write $$\lim_{x\to 8}\frac{\root{3}\of{x} - 2}{\root{3}\of{3x+3}-3}=\lim_{x\to 8}\frac{\frac{\root{3}\of{x} - 2}{x-8}}{\frac{\root{3}\of{3x+3}-3}{x-8}}=\frac{\lim_{x\to 8}\frac{\root{3}\of{x} - 2}{x-8}}{\lim_{x\to 8}\frac{\root{3}\of{3x+3}-3}{x-8}}$$
You then have the limit as the ratio of the derivatives (you derived l'Hospital rule)
A: You want to calculate the limit with the help of a composite function:
$$\lim_{x\rightarrow 8}\frac{\root{3}\of{x} - 2}{\root{3}\of{3x+3}-3}.$$
There are two problems in your solution:
1) You actually implied $f(x)\equiv h(x)$. So, $h(x) = \root{3}\of{3x+3}, g(x) = \frac{x-3}{3}$, then $h(g(x))=\sqrt[3]{x}$ and $h(g(8)) = 2$.
2) The formal definition of the derivative of a composite function is: $\lim_\limits{x\rightarrow a} \frac{h(g(x)) - h(g(a))}{x - a}$. So:
$$\lim_{x\rightarrow 8} \frac{h(g(x)) - h(g(8))}{h(x) - h(8)}=\\
\lim_{x\rightarrow 8} \frac{h(g(x)) - h(g(8))}{x - 8}\cdot \lim_{x\rightarrow 8} \frac{x - 8}{h(x) - h(8)}=\\
(\sqrt[3]{x})'_{x=8}\cdot \left((\sqrt[3]{3x+3})'_{x=8}\right)^{-1} =\\
\frac{1}{3\sqrt[3]{8^2}}\cdot \left(\frac{3}{3\sqrt[3]{(3\cdot 8+3)^2}}\right)^{-1}=\\
\frac1{12}\cdot 9=\frac34.$$
