Exchange a limit with a $\limsup$ or $\liminf$

Question

Let $f_{n}:\,\mathbb{R}\rightarrow\mathbb{R}$ a sequence of $C^{1}\left(\mathbb{R}\right)$ functions such that $$\left|f_{n}\left(x\right)\right|\leq M,\ \forall x\in\mathbb{R},\,\forall n\in\mathbb{N}$$and assume that $$\lim_{n\rightarrow\infty}f_{n}\left(x\right)=f\left(x\right)$$ for all $x$ in $\mathbb{R}$. Can I conclude that $$\lim_{n\rightarrow\infty}\limsup_{h\rightarrow0}\frac{f_{n}\left(x+h\right)-f_{n}\left(x\right)}{h}=\limsup_{h\rightarrow0}\lim_{n\rightarrow\infty}\frac{f_{n}\left(x+h\right)-f_{n}\left(x\right)}{h}$$ $$=\limsup_{h\rightarrow0}\frac{f\left(x+h\right)-f\left(x\right)}{h}?$$ I know that there are some theorems that allow to switch two limits, like the dominated convergence theorem, but what can I say in this situation? I have not more information about these functions. Note that I'm interested also in the case that the limit is not convergent. In other words: is it true that $$\lim_{n\rightarrow\infty}\limsup_{h\rightarrow0}\frac{f_{n}\left(x+h\right)-f_{n}\left(x\right)}{h}=L\Leftrightarrow\limsup_{h\rightarrow0}\frac{f\left(x+h\right)-f\left(x\right)}{h}=L,\,L\in\mathbb{R}$$ or $$\lim_{n\rightarrow\infty}\limsup_{h\rightarrow0}\frac{f_{n}\left(x+h\right)-f_{n}\left(x\right)}{h}\text{ does not exist}\Leftrightarrow\limsup_{h\rightarrow0}\frac{f\left(x+h\right)-f\left(x\right)}{h}\text{ does not exist}$$ or $$\lim_{n\rightarrow\infty}\limsup_{h\rightarrow0}\frac{f_{n}\left(x+h\right)-f_{n}\left(x\right)}{h}=\infty\,(-\infty)\Leftrightarrow\limsup_{h\rightarrow0}\frac{f\left(x+h\right)-f\left(x\right)}{h}=\infty\,(-\infty)?$$

Answer 1

No, this is not correct. For example, set $f_n(x)=\frac{1}{n}\arctan(nx)$. Then $f\equiv 0$, and at $x=0$, we have that $$\lim_{n\to\infty}\limsup_{h\to 0}\frac{f_{n}(x+h)-f_{n}(x)}{h}=\lim_{n\to\infty}\limsup_{h\to 0}\frac{\arctan(nh)}{nh}=1,$$ while $$\limsup_{h\rightarrow0}\frac{f\left(x+h\right)-f\left(x\right)}{h}=0.$$