real analysis - Function that is not uniformly differentiable

Define $f: A \to \mathbb R$ ($f$ differentiable)to be uniformly differentiable if and only if for $\varepsilon >0$ there exists $\delta >0$ such that

$$ |h| < \delta \implies \left| {f(x + h) - f(x) \over h} -f'(x) \right | < \varepsilon$$

I am looking for an example of $f$ that is not uniformly differentiable. My idea was to choose $f$ with $f'$ sufficiently steep. For example, $f(x) = {1 \over x}$. But on $[1, \infty)$

$$ \left| {f(x + h) - f(x) \over h} -f'(x) \right | = \left| {h \over x^2 (x+h)} \right | \ge \left| {h \over x+h} \right |$$

and I don't know how to use this. On $(0,1)$ I similarly can't find a lower bound on this either. Maybe $1/x$ is in fact uniformly differentiable but I coudln't bound it from above either.

My questions are: Is $1/x$ uniformly differentiable or not and how to show it. Also can you please give me an example of $f$ that is not uniformly differentiable.

Answer

Let $f(x) = 1/x$. As you showed,

$$
\left| \frac{f(x + h) - f(x)}{h} - f'(x) \right|
= \left| \frac{h}{(x+h)x^2} \right| \\
$$
For any $\delta > 0$, you can just let $h = x = \frac{1}{M}$, for some sufficiently large $M$.Then
$$
\left| \frac{f(x + h) - f(x)}{h} - f'(x) \right|
= \left| \frac{1\; / \; M}{(2 \; / \; M) (1 \; / \; M)^2} \right|
= \frac{M^2}{2}
$$

so this function is certainly not uniformly differentiable.
The point is that the condition for uniform differentiability requires the same value of $\boldsymbol\delta$ for all $\boldsymbol{x, h}$, i.e. $\delta$ does not depend on $x$ and $h$.