L'Hopital's rule and limiting variables I'm working some problems from a calculus text and came across this question:
If $f(x)$ is a function that's differentiable everywhere, what is the value of the limit $$\lim\limits_{h \to 0}\frac{f(x+3h^2)-f(x-h^2)}{2h^2}$$
I know that it is necessary to use L'Hopital's and at first I was taking derivatives with respect to x. However, the book's answer arrives at the solution by differentiating with respect to h. 
I'm confused as to why differentiating with respect to h is the proper way to approach this problem, I was under the impression that $f$ is a function of x. I did some research and the term "limiting variable" is used a lot, but I'm still a little mixed up.
 A: L'Hospital's rule does not work here and we can directly work using the substitution $t = h^{2}$ as follows:
\begin{align}
L &= \lim_{h \to 0}\frac{f(x + 3h^{2}) - f(x - h^{2})}{2h^{2}}\notag\\
&= \lim_{t \to 0^{+}}\frac{f(x + 3t) - f(x - t)}{2t}\notag\\
&= \lim_{t \to 0^{+}}\frac{f(x + 3t) - f(x) + f(x) - f(x - t)}{2t}\notag\\
&= \lim_{t \to 0^{+}}\frac{f(x + 3t) - f(x)}{2t} + \frac{f(x) - f(x - t)}{2t}\notag\\
&= \lim_{t \to 0^{+}}\frac{f(x + 3t) - f(x)}{3t}\cdot\frac{3}{2} + \frac{1}{2}\cdot\frac{f(x - t) - f(x)}{(-t)}\notag\\
&= \frac{3}{2}\cdot\lim_{u \to 0^{+}}\frac{f(x + u) - f(x)}{u} + \frac{1}{2}\cdot\lim_{v \to 0^{-}}\frac{f(x + v) - f(x)}{v}\text{ (putting } u = 3t, v = -t)\notag\\
&= \frac{3}{2}\cdot f'(x) + \frac{1}{2}\cdot f'(x)\notag\\
&= 2f'(x)\notag
\end{align}
A: Let $$L = \lim\limits_{h \to 0}\frac{f(x+3h^2)-f(x-h^2)}{2h^2}$$
Now When $h\rightarrow 0\;,$ Then limit $\displaystyle L$  are in $\displaystyle  \frac{0}{0}$ form and Here $h$ is a variable and $x$ is Constant
So when we apply $\bf{D,LHopital\; Rule}\;,$ We Differentiate it  with respect to $h$
So $$\displaystyle L = \lim_{h\rightarrow 0}\frac{f'(x+3h^2)\cdot 6h-f'(x-h^2)\cdot (-2h)}{4h}$$
So $$\displaystyle  = \lim_{h\rightarrow 0}\frac{f'(x+3h^2)\cdot 6-f'(x-h^2)\cdot (-2)}{4}$$
So We get $$\displaystyle L = \frac{6f'(x)+2f'(x)}{4} = 2f'(x)$$
