<s>[INST] Find $\lim\limits_{h\to 0}\frac{f(2-h^2)-f(2)}{h^2}$ let $f(x)= (x^3+2x)[\frac{x}{2}]$ 
[x]:floor function 
then :
$$\lim\limits_{h\to 0}\frac{f(2-h^2)-f(2)}{h^2}=?$$
My try :
$$\lim\limits_{h\to 0}\frac{f(2-h)-f(2)}{h}=f'(2)$$
$$\lim\limits_{x\to 2}\frac{(x^3+2x)[\frac{x}{2}]-f(2)}{x-2}=f'(2)$$
$$\lim\limits_{x\to 2}\frac{(x^3+2x)[\frac{x}{2}]-12}{x-2}=f'(2)$$
Now what ?
 [/INST] Observe that as $h\rightarrow 0$, $h^2\rightarrow 0^+$ since $h^2$ is always nonnegative.  Therefore, we can write:
$$
\lim_{h\rightarrow 0}\frac{f(2-h^2)-f(2)}{h^2}=\lim_{k\rightarrow 0^+}\frac{f(2-k)-f(2)}{k}
$$
by setting $k=h^2$.  Then, by substituting the formula for $f$, we get
\begin{align*}
\lim_{k\rightarrow 0^+}\frac{f(2-k)-f(2)}{k}&=\lim_{k\rightarrow 0^+}\frac{((2-k)^3+2(2-k))\left\lfloor\frac{2-k}{2}\right\rfloor-(2^3+4)\left\lfloor\frac{2}{2}\right\rfloor}{k}\\
&=\lim_{k\rightarrow 0^+}\frac{(-k^3+6k^2-14k+12)\left\lfloor\frac{2-k}{2}\right\rfloor-12}{k}
\end{align*}
Since, when $k$ is slightly larger than $0$, $2-k$ is slightly less than $2$, $\frac{2-k}{2}$ is slightly less than $1$, so its floor is $0$.  This implies that we're looking at 
$$
\lim_{k\rightarrow0^+}\frac{-12}{k}
$$
which does not exist.
On the other hand,
$$
\lim_{h\rightarrow 0}\frac{f(2+h^2)-f(2)}{h^2}
$$ 
does exist and the limit is $14$.
</s>