$n$:th roots of monic polynomials (of order $n$) have the curious property:
$$\lim_{x\to \infty}\sqrt[n]{x^n+c_{n-1}x^{n-1}+\cdots+c_0} = x + k$$
They have linear asymptotes.
Now, enter Huber loss function.
$$L_\delta(a)=\cases{\frac 1 2 a^2\hspace{2cm} \text{ for } |a|<\delta,\\\delta (|a|-\frac 1 2 \delta)\hspace{0.75cm} \text{for } |a|>\delta}$$
It is quadratic for small arguments, but linear once we get outside of $\delta$. Will functions of the form prescribed above be able to approximate our function arbitrarily well? If not, why not?
Just to complement the great answer below, I made some numeric experiments. With numeric optimization methods for example in GNU Octave we find approximations with polynomials of order 2,4 and 6:
Errors to the left. Curve shape to the right.
I think the second order parabola stops at $x=1$ and there the line starts.
Let me reformulate your assertion under the precise form :
$$\lim_{x\to \infty}\left(\sqrt[n]{x^n+q(x)}-x\right)=\dfrac{1}{n}c_{n-1} \ \ \ \text{where} \ \ \ q(x):=c_{n-1}x^{n-1}+\cdots+c_0 \tag{1}$$
and prove it. Indeed :
$$\sqrt[n]{x^n+q(x)}=\sqrt[n]{x^n\left(1+\dfrac{q(x)}{x^n}\right)}=x\sqrt[n]{1+\dfrac{q(x)}{x^n}}.\tag{2}$$
Now, let us use the first two terms of the classical "generalized Newton binomial expansion" :
$$(1+X)^{a}=1+aX+\dfrac{a(a-1)}{2}X^2+... \text{with} \ \ a:=\dfrac{1}{n}, \ \ X:=\dfrac{q(x)}{x^n}$$
expression (2) becomes :
$$x \left(1+\dfrac{1}{n}\dfrac{q(x)}{x^n}\right)=x+\dfrac{1}{n}\dfrac{c_{n-1}x^{n-1}+...}{x^{n-1}} \approx x+\dfrac{1}{n}c_{n-1}\tag{3}$$
(this approximation being valid for $x \to \infty$).
Plugging (3) into (2), (1) is established.
Remark : Our derivation, strictly speaking, is valid "for $x$ sufficiently large", i.e., for $x$ such that the expression under the radical symbol is always positive. Otherwise, we should have taken absolute values under the radical sign in order to prevent ambiguous cases.
Besides, I don't see any connection with Huber loss function.