Is there a function $h$ and a constant $c$ to satisfy $h(x-c) \neq O(h(x))$ Is there a function $h$ and a constant $c$ to satisfy $h(x-c) \neq O(h(x))$?

My attempt:
This claim isn't true. we shall show an equivalent claim to the negation of our claim. We will show that  for each function $h(x)$ and a constant $c$, from a certain $x$, we get $h(x-c)\le c_1\cdot h(x)$. Now I don't know how to proceed, however, my intuition says that $h(x-c)$ is the same function $h(x)$ but when you move it horizontally $c$, so by that fact we can pick $c_1$ such that will be big enough to overcome $h(x-c)$, because we have a bigger slope of $h(x)$. Therefore, we get that: $$h(x-c)\le c_1\cdot h(x)$$
I am not sure that this is formal, and probably I am missing details, so I will be glad to get some help. Thanks!
 A: Do we allow $c<0$? In that case maybe pick $h(x) = e^{x^2}$, then
$$ h(x-c) = e^{x^2 -2cx + c^2} > e^{2|c|x} e^{x^2} \neq O(h(x)).  $$
In fact, for $c>0$ simply pick $h(x) = e^{-x^2}$, then
$$ h(x-c) = e^{-x^2 +2cx - c^2} = e^{-c^2} e^{2cx} e^{-x^2} \neq O(h(x)).$$
To see why this is true, recall that $f(x) = O(g(x))$ iff there are constants $x_0$ and $M$ such that for all $x\geq x_0$ we have $f(x) \leq M g(x)$. Suppose for contradiction that $h(x-c) = O(h(x))$, and let $x_0$ and $M$ be the corresponding constants. Then for all $x\geq x_0$ we have
$$  e^{-c^2 }e^{2cx} e^{ -x^2}\leq M e^{-x^2}. $$
Cancelling $e^{-x^2}$ from both sides, we now have constants $x_0$ and $M$ such that $x\geq x_0$ implies $e^{-c^2 }e^{2xc} \leq M$, but the left hand side of this goes to infinity! So this is a contradiction, and indeed we have $h(x-c) \neq O(h(x))$.
I think some of the confusion may come from the fact that $h(x)$ and $h(x-c)$ both go to zero as $x\to \infty$. But in this example $h(x-c)$ goes to infinity much slower, as there is an additional $e^{2cx}$ term which will be quite large.
Another counter example: This is kinda fun so I thought of another example of $h(x)$ and $c$ with $h(x-c) \neq O(h(x))$. Let $h(x)$ defined as
$$h(x) = \begin{cases}0 &\text{ if }x\in \mathbb{Q},\\ 
 x& \text{otherwise}.\end{cases} .$$
If $c$ is any irrational number then for all rational $x$ we have $h(x-c)=x-c$ while $h(x) = 0$. So, fixing such an irrational $c$, there are rational $x > c$ for which $h(x-c)-h(x)$ exceeds any bound.
Similarly, we can use e.g. $h(x)= 1/(\sin (\pi x) + 1.1)$ to achieve a similar effect.
