For all $\epsilon >0 $, exists $l(\epsilon)$ such that $\forall x \in \mathbb{R}$ and $\phi \in (0, l(\epsilon)), |f(x+\phi) -f(x)| < \epsilon$ Problem
Let $f:\mathbb{R} \rightarrow \mathbb{R}$ differentiable such that $|f'(x)|<K$. Prove for all $\epsilon >0 $, exists $l(\epsilon)$ such that $\forall x \in \mathbb{R}$ and $\phi \in (0, l(\epsilon))$
$$|f(x+\phi) -f(x)| < \epsilon.$$
Attempt
Let $\epsilon >0$. $\forall x \in \mathbb{R}$, $$|\frac{f(x+h)-f(x)}{h}-f'(x)| < \epsilon \Rightarrow |f(x+h)-f(x)-hf'(x)| < |h|\epsilon$$
Then, $|f(x+h)-f(x)|-|h|K \leq |f(x+h)-f(x)-hf'(x)|$,
after, $|f(x+h)-f(x)| < |h|(\epsilon+K)$
But I have problems here, because $h$ is not fixed, if it was we would take $h=\frac{\epsilon}{\epsilon +k}$, could you help me?
 A: 
Problem Let $f:\mathbb{R} \rightarrow \mathbb{R}$ differentiable such that $|f'(x)|<K$. Porve for all $\epsilon >0 $, exists $l(\epsilon)$ such that $\forall x \in \mathbb{R}$ and $\phi \in (0, l(\epsilon))$


$$|f(x+\phi) -f(x)| < \epsilon.$$


Attempt Let $\epsilon >0$. $\forall x \in \mathbb{R}$, $$|\frac{f(x+h)-f(x)}{h}-f'(x)| < \epsilon \Rightarrow |f(x+h)-f(x)-hf'(x)| < |h|\epsilon$$

$$\forall \epsilon>0,\forall x,  \exists\delta_\epsilon (x)>0, 0<|h-0|<\delta_\epsilon (x)\Rightarrow |\frac{f(x+h)-f(x)}{h}-f'(x)|<\epsilon$$
$(1)$ When you use the condition that the derivative $f'(x)$ exists pointwisely, your $\delta_\epsilon (x)$ depends on $x$, but you didn't write anything about your $\delta_\epsilon (x)$ and how is it related to $l(\epsilon)$.
$(2)$ Your final goal is to show $f(x)$ is uniformly continuous, which means you need to find $l(\epsilon)$, which is independent of $x$. So you have to revise all the work afterwards.
The proof is to use Mean Value Theorem.
Since $f'(x)$ is bounded s.t.
$$|f'(x)|\leq K\quad \forall x\in\mathbb{R}$$
So by Mean Value Theorem we have
$$|f(x)-f(y)|=|f'(\xi)|\cdot |x-y|\leq K|x-y|\quad \forall x,y\in\mathbb{R}$$
So take $l(\epsilon)=\frac{\epsilon}{K}$,then you can show $f(x)$ is uniformly continuous on $\mathbb{R}$.
A: Since $f$ is differentiable we have
$|f(x+h)-f(x)| = |h||f'(\alpha)|$ for some $\alpha \in (x, x+h)$ by mean-value theorem, which means $|f(x+h)-f(x)| \le K |h|$. For given $\epsilon >0$ let $l(\epsilon) = \frac{\epsilon}{K}$. Then for any $\phi \in (0,l(\epsilon))$, we have
$|f(x+\phi)-f(x)| = |\phi||f'(\beta)| \le \frac{\epsilon}{K} \,K =\epsilon$.
