Is this an immediate consequence of the Straddle Lemma? As main book, I'm using Bartle, Introduction to Real Analysis (2011 4 ed). Exercise 17 Section 6.1, p 171 asks you to prove the Straddle Lemma:



*Let $f:I\rightarrow\mathbb R$ be differentiable at $c\in I$. Establish the Straddle Lemma: Given $\varepsilon >0 $ there exists $\delta >0$ such that if $u, v\in I$ satisfy $c-\delta \lt u\leq c\leq v\lt c+\delta$ then we have $|f(v)-f(u)-(v-u)f'(c)|\leq \varepsilon (v-u).$
[Hint: The $\delta$ is given by Definition 6.1.1. Subtract
and add the term $f(c) - cf'(c)$ on the left side and use the Triangle Inequality.]


It is sufficient to know the formal definition of derivative to prove this result. Later, in exercise 18 of section 6.2, p 180 (that is based on the Mean Value Theorem and some of its consequences) asks



*Let $I:=[a,b]$ and let $f:I\rightarrow\mathbb R$ be differentiable at $c\in I$. Show that for every $\varepsilon >0 $ there exists $\delta >0$ such that if $\;0<|x-y|\lt \delta$ and $a \leq x \leq c \leq y \leq b$, then
$$\left|\frac {f(x)-f(y)}{x-y}-f'(c)\right|\lt \varepsilon$$

Is this a direct consequence of the Straddle Lemma?
I didn't get to anything keeping distance from the Straddle Lemma and trying to use the Mean Value Theorem. I'm unsure about using the Straddle Lemma firstly because it gives me a not strict inequality and the second exercise has a strict inequality, and secondly because I'm not using any "tools" provided from the respective section. Is the Straddle Lemma useful? If so, how can I end up with a strict inequality? If not, any hints would be appreciated. Thank you in advance.
 A: You don't need Straddle Lemma.


*Given $ε>0$, let $δ =δ(ε)$ be as in Definition 6.1.1 on p 161 (see bottom), and let $x<c<y$ be such
that $0<|x−y|<δ$. The trick here is to rewrite $\left|\dfrac{f(x)-f(y)}{x-y}-f'(c)\right|$, and exploit that $\left(\dfrac{\color{red}{x-c}}{x-y}+\dfrac{\color{skyblue}{c-y}}{x-y}\right) = 1$.

$\begin{align} \left|\frac {f(x)-f(y)}{x-y}-f'(c)\right| & = \left|\frac{f(x)\color{limegreen}{-f(c)}}{x-y}+\frac{\color{limegreen}{f(c)}-f(y)}{x-y}-\left(\frac{\color{red}{x-c}}{x-y}+\frac{\color{skyblue}{c-y}}{x-y}\right)f'(c)\right| \\
&\leq\left|\frac{f(x)-\color{limegreen}{f(c)}}{x-y}-\frac{\color{red}{x-c}}{x-y}f'(c)\right|\color{limegreen}{+}\left|\frac{\color{limegreen}{f(c)}-f(y)}{x-y}-\frac{\color{skyblue}{c-y}}{x-y}f'(c)\right|\\
& \text{Notice that both} \frac{\color{red}{x-c}}{x-y}, \frac{\color{skyblue}{c-y}}{x-y} \text{ are positive. And  factor them out }. \\ 
& = (\frac{\color{red}{x-c}}{x-y})\left|\frac{f(x) - f(c)}{\color{red}{x-c}} - f'(c) \right| +(\frac{\color{skyblue}{c-y}}{x-y})\left|\frac{f(y) - f(c)}{\color{skyblue}{c-y}} - f'(c)\right|  \\
& < (\frac{x-c}{x-y} +  \frac{c-y}{x-y})ε \\
& \qquad =ε \end{align}$
Note that if one  of $x$ and $y$ (but not both) equals $ε$, the conclusion still holds.


