Continuously differentiable function iff $|f(x + h) - f(x + t) - l(h - t)| \leq \epsilon |h-t|$ I am doing some practice tests for my analysis class, and in one of them I stumbled upon the following question:
Prove: $f: \mathbb{R} \rightarrow \mathbb{R}$ is continuously differentiable iff for every $x \in \mathbb{R}$ there exists a $l \in \mathbb{R}$ such that for every $\epsilon > 0$ there exists a $\delta > 0$ such that for every $h,t \in B(0, \delta)$:
$|f(x + h) - f(x + t) - l(h - t)| \leq \epsilon |h-t|$
I was hoping someone could walk me through this proof. Whatever I try, I can't  get to this form by using the $\epsilon-\delta$ definition of continuity and differentiability. I have also tried continuously differentiable $\rightarrow$ Lipschitz, but that did not get me any further. 
Thanks in advance!
 A: If you take $t=0$ you get $|f(x+h)-f(x)-lh|\le\epsilon|h|$, which is exactly the definition of differentiability. So you know that $l=f^\prime(x)$.
Not you need to prove that $f^\prime$ is continuous, that is that for every $\epsilon$ and $x$ there is $\delta$ such that $| f^\prime(x+h)-f^\prime(x)|\le\epsilon$ for $|h|\le \delta$. Assume by contradiction that this is not true, then you find $\epsilon$ such that for every $\delta$ there is $|h|\le\delta$ with $| f^\prime(x+h)-f^\prime(x)|>\epsilon$. Take $\delta=\frac1n$ and $|h_n|\le\frac1n$ with $| f^\prime(x+h_n)-f^\prime(x)|>\epsilon$. Since $$\left|\lim_{t\to 0}\frac{f(x+h_n+t)-f(x+h_n)}{t}-f'(x)\right|>\epsilon,$$ you find
$\left|\frac{f(x+h_n+t_n)-f(x+h_n)}{t_n}-f'(x)\right|>\epsilon$ for some $t_n$ as small as you want, that is $\left|f(x+h_n+t_n)-f(x+h_n)-f'(x)t_n\right|>\epsilon |t_n|$, which contradicts your hypothesis.
A: Hint. By $|f(x_0 + h) - f(x_0 + t) - l_0(h - t)|\leq \epsilon |h-t|,$ we have
$$ 
\Big|\big(f(x_0 + h)-f(x_0)-l_0\cdot h\big) - \big(f(x_0 + t)-f(x_0) - l_0\cdot t\big)\Big|
\leq 
\epsilon |h-t|
$$
Since $t$ and $h$ are arbitrary in $B(0,\delta)$ we can make $t = 0$ and obtain
$$ 
(\ast)\quad\frac{|f(x_0 + h)-f(x_0)-l_0\cdot h|}{|h|} 
\leq 
\epsilon. 
$$
By doing $\epsilon \to 0$ we derive the derivative of $f(x)$ in $x=x_0$: $f'(x_0)=l_0$. Now to get the continuity of $f'(x)$ in $x=x_0$ note that for all $t\in B(0,\delta)$ we have
\begin{align} 
\left|
f'(x_0+h)-f'(x_0)
\right|
=&
\frac{1}{|t|}
\cdot
\left|
f'(x_0+h)\cdot t-f'(x_0)\cdot t
\right|
\\
=&\frac{1}{|t|}
\cdot
\left|
f'(x_0+h)\cdot t-f(x_0+h+t)+f(x_0+h)
\right.
\\
&
\left.
\qquad + \;\;f(x_0+h+t)-f(x_0+h)
-f'(x_0)\cdot t
\right|
\\
\leq &
\frac{1}{|t|}
\cdot
\left|
f(x_0+h+t)-f(x_0+h)-f'(x_0+h)\cdot t
\right|
\\
&
\quad + \frac{1}{|t|}
\cdot
\left|
f(x_0+h+t)-f(x_0+h) -f'(x_0)\cdot t
\right|
\end{align}
By $(\ast)$, $| f'(x_0+h)-f'(x_0)|\leq \epsilon + \epsilon$.
