Distance between polynomials $x^n$ and $x^n-x^{n-1}$ For polynomials $p(x)=x^n$ and $q(x)=x^n-x^{n-1}$, let $z(x)=\min_{a\in\mathbb{R}}\sqrt{(x-a)^2+\big(p(x)-q(a)\big)^2}$ be the distance between the point $\langle x,p(x)\rangle$ and the nearest point on the graph of $q(\cdot)$.  I've noticed that it looks like this distance approaches a finite limit; i.e., $\displaystyle\lim_{x\to\infty}z(x)=c$ for some $c$. Is this true and, if so, how can I find $c$? https://www.desmos.com/calculator/foqvlfqyhh
 A: While finding the exact value of the limit is a little bit trickier, it's straightforward to show that such a limit exists: we can show that the value $x'$ with $q(x')=p(x)$ is within a bounded distance of $x$, and this clearly provides an upper bound on the distance-between-graphs requested here.  Indeed, just let $w=x+1$; then $q(w)=(x+1)^n-(x+1)^{n-1}$ $=x^n+nx^{n-1}+O(x^{n-2})-(x^{n-1}+O(x^{n-2}))$ $=x^n+(n-1)x^{n-1}+O(x^{n-2})$.  Since $n\gt 1$, this will be $\gt x^n$ for all sufficiently large $x$.  But since $q(x)\lt p(x)$ and $q(w)\gt p(x)$, there's some $\zeta\in[x,w]$ with $q(\zeta)=p(x)$. But this means that $z(x) \leq \sqrt{(\zeta-x)^2+(q(\zeta)-p(x))^2}$ $=\sqrt{(\zeta-x)^2}=|\zeta-x|\leq|w-x|=1$ and so $z(x)\leq 1$ for all sufficiently large $x$.
In fact, if we set $w=x+t$ here (with $t$ to be determined) and run the same analysis, we get $q(w) = x^n+ntx^{n-1}-x^{n-1}+O(x^{n-2})$, so for $t\approx \frac1n$ (plus smaller terms) we already have $q(x+t)\approx p(x)$.  I suspect that you'll find the exact value of your constant $c$ is comparable to the horizontal distance, $\frac1n+O(n^{-2})$, but it might be hard to get a precise handle on the higher-order terms.
