Vrftjkry

I have to calculate the folowing:

$$lim_{xrightarrow 8}frac{root{3}of{x} - 2}{root{3}of{3x+3}-3}$$

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

$$f'(a) = lim_{xrightarrow a} frac{f(x) - f(a)}{x - a}$$

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:

$$
lim_{xrightarrow 8} frac{h(g(x)) - h(g(8))}{h(x) - h(8)}
$$ (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 $xrightarrow 8$? 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

$$
frac{dc}{dx} = frac{1}{3root{3}of{(3x+3)^2}} \
frac{dh}{dx} = frac{3}{3root{3}of{(3x+3)^2}} \
Rightarrow frac{dc}{dh} = frac{1}{3}
$$

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 $frac{1}{3root{3}of{x^2}}$. This leads to the correct limit, as showed in farruhota's answer.

You want to calculate the limit with the help of a composite function:
$$lim_{xrightarrow 8}frac{root{3}of{x} - 2}{root{3}of{3x+3}-3}.$$

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: $lim_limits{xrightarrow a} frac{h(g(x)) - h(g(a))}{x - a}$. So:
$$lim_{xrightarrow 8} frac{h(g(x)) - h(g(8))}{h(x) - h(8)}=\
lim_{xrightarrow 8} frac{h(g(x)) - h(g(8))}{x - 8}cdot lim_{xrightarrow 8} frac{x - 8}{h(x) - h(8)}=\
(sqrt[3]{x})'_{x=8}cdot left((sqrt[3]{3x+3})'_{x=8}right)^{-1} =\
frac{1}{3sqrt[3]{8^2}}cdot left(frac{3}{3sqrt[3]{(3cdot 8+3)^2}}right)^{-1}=\
frac1{12}cdot 9=frac34.$$

We don't need composite function, indeed we can use that

$$lim_{xrightarrow 8}frac{root{3}of{x} - 2}{root{3}of{3x+3}-3}
=lim_{xrightarrow 8}frac{root{3}of{x} - 2}{x-8}frac{x-8}{root{3}of{3x+3}-3}$$

and use derivative definition for $f(x)=root{3}of{x}$ and $g(x)=root{3}of{3x+3}$.

Assuming that the derivative of the numerator and denominator exist and are finite, then we can write $$lim_{xto 8}frac{root{3}of{x} - 2}{root{3}of{3x+3}-3}=lim_{xto 8}frac{frac{root{3}of{x} - 2}{x-8}}{frac{root{3}of{3x+3}-3}{x-8}}=frac{lim_{xto 8}frac{root{3}of{x} - 2}{x-8}}{lim_{xto 8}frac{root{3}of{3x+3}-3}{x-8}}$$
You then have the limit as the ratio of the derivatives (you derived l'Hospital rule)