Verify: Uniformly differentiable functions have continuous derivatives Let $f$ be uniformly differentiable if for all $\varepsilon > 0$, there exists $\delta > 0$ such that for all $x, y \in \mathbb R$,
$$\left \lvert \frac {f(x)-f(y)}{x-y} - f'(y) \right \rvert  < \varepsilon
$$ when $|x-y| < \delta$.  Prove that if $f$ is uniformly differentiable, then $f'$ is continuous.
Note: Proofs are available; this question is to verify and critique this proof.
Proof: Suppose that $f$ is uniformly differentiable.  Observe that for all $|x-y| < \delta$, the symmetric inequality $$\left \lvert \frac {f(x)-f(y)}{x-y} - f'(x) \right \rvert  < \varepsilon
$$ also holds, since $\frac {f(x)-f(y)}{x-y} =  \frac {f(y)-f(x)}{y-x}.$  Fix an arbitrary $x \in \mathbb R$ and choose $\delta$ such that
$$\left \lvert \frac {f(x+h)-f(x)}{h} - f'(x) \right \rvert  < \frac \varepsilon 3\\
$$
$$
\left \lvert \frac {f(x)-f(x-h)}{h} - f'(x) \right \rvert  <  \frac \varepsilon 3 $$ for $|h| < \delta$.  Then, by the triangle inequality,
$$
\left \lvert \frac {f(x+h)-f(x)}{h} - \frac{f(x)-f(x-h)}{h} \right \rvert < \frac 2 3 \varepsilon
$$ for all $|h| < \delta$.
This in turn implies that
$$
| f'(x+h) - f'(x) | \leq \frac 2 3 \varepsilon < \varepsilon
$$ for all $|h| < \delta$, since if $|a_n - b_n| < k$ for all $n$, $|\lim a_n - \lim b_n| \leq k$.  This proves that $f'$ is continuous.
Questions: Is this proof correct, rigorous, and well-written? Did I justify each step sufficiently ? How can it be improved?
 A: All of your steps look good to me except your "This in turn implies that..." sentence. You've written
$$\left \lvert \frac {f(x+h)-f(x)}{h} - \frac{f(x)-f(x-h)}{h} \right \rvert < \frac 2 3 \varepsilon$$
and then you're trying to somehow pass to a limit to conclude
$$| f'(x+h) - f'(x) | \leq \frac 2 3 \varepsilon < \varepsilon$$
but you aren't being rigorous here. What limit are you taking? If you take the limit of $\frac {f(x+h)-f(x)}{h} - \frac{f(x)-f(x-h)}{h}$ as $h \to 0$ then you'll just get $f'(x) - f'(x)$ which is $0$.
The intuition behind your proof strategy is fine, but right now you're trying to take a limit where "some of the $h$s get limited away and others remain", which is not a valid operation. We can save the proof by introducing a new variable which I'll call $s$. Then for $|h|, |s| < \frac{\delta}{2}$ you can use your same argument to get
$$\left \lvert \frac {f(x+h)-f(x)}{h} - \frac{f(x-s+h)-f(x-s)}{h} \right \rvert < \frac 2 3 \varepsilon$$
and now you can take the limit as $h \to 0$ to get
$$| f'(x) - f'(x-s) | \leq \frac 2 3 \varepsilon < \varepsilon.$$
Then you can finish the proof the same way you're currently doing.
Does that make sense? Basically my complaint is "you can't take a limit as some of your $h$s go to 0 while others stick around", and in general that mistake could lead you to "proofs" of totally false statements, but in this case it's easy to fix just by making the $h$s that shouldn't get limitified into a separate variable.
Side note: Martin R's comment is how I would have proved this statement. If you're handing this in somewhere then I'd recommend 1) understanding his idea, then 2) still fixing+using your own proof instead.
