Show that when $f,g:\mathbb{R} \rightarrow \mathbb{R}$,and for every $\varepsilon >0$, there exist $\delta=\delta(\varepsilon)>0$, and $|h|<\delta$ guarantees $|\frac{f(x+h)-f(x)}{h}-g(x)|<\varepsilon.$
Show that that $f^{'}$ exist and is continuous for every $x \in \mathbb{R}$.
My idea: Let $\epsilon>0$, there exist a $\delta>0 $ such that if $y\in(x-\delta,x+\delta)$ and $|h|<\delta$. then
$|\frac{f(y+h)-f(y)}{h}-g(y)|<\varepsilon \implies g(y)-\varepsilon<\frac{f(y+h)-f(y)}{h}<g(y)+\varepsilon.$
Now, $\limsup_{h \rightarrow 0}g(y)-\varepsilon<\limsup_{h \rightarrow 0} \frac{f(y+h)-f(y)}{h}< \limsup_{h \rightarrow 0} g(y)+\varepsilon$.
This implies $g(y)-\varepsilon<\limsup_{h \rightarrow 0} \frac{f(y+h)-f(y)}{h}< g(y)+\varepsilon$.
This implies $|f^{'}(y)-g(y)|<\varepsilon$ for all $y \in (x-\delta,x +\delta)$.
I observe that $f^{'}$ exist when $g(y)$ exist, but how I guarantee that $g(y)$ exist here.
Can anyone suggest me some hints for this question?
By definition of derivative, it is clear that $$f’(x)=g(x).$$ Furthermore $f(x)$ is continuous and so is $$f_n(x):=n\left(f\left(x+\frac 1 n\right)-f(x)\right),n\in {\mathbb N}.$$
Referring to the given condition, one has, $\forall \epsilon>0,\exists \delta$ such that $|h|<\delta$ implies $$\left|\frac{f(x+h)-f(x)}h-g(x)\right|<\epsilon,$$ thus $$|f_n(x)-g(x)|<\epsilon,$$ provided $n>\left[\frac 1{\delta}\right]+1,$ independent of $x$. It follows that $f_n(x)$ converges to $g(x)$ uniformly, hence $g(x)$ is continuous.