In a real normed linear space if $||x||=||y||$ implies $\lim_{n \to \infty} ||x+ny||-||nx+y||=0$ , then the norm comes from an inner-product space? $(V,\|\cdot|)$ be a real  normed linear space such that
$\|x\|=\|y\|$ implies $\lim\limits_{n\to\infty} \|x+ny\|-\|nx+y\|=0$, then is it true that the norm comes from an 
inner-product space ? 
 A: Since no one provided the answer, I write mine. So, it seems the following. 
Parallelogram law implies that it suffices to verify the claim for two-dimensional vector spaces.
So, let on the plane $\Bbb R^2$ is defined a norm $\|\cdot\|$ satisfying the proposed equality. Let $K=\{x\in\Bbb R^2:\|x\|\le 1\}$ be the unit ball of the norm $\|\cdot\|$. Since the norm $\|\cdot\|$ is a continuous function, $K$ is a closed subset of the plane. Since $K$ is a bounded set, it is compact. Since $\|\cdot\|$ is a norm, $K$ is convex. As $\|\cdot\|$ we denote the standard norm on the plane. For each angle $\alpha\in\Bbb T=\{z\in\Bbb R^2:|z|=1\}$ put  $f(\alpha)=\sup\{\lambda>0:\lambda\alpha\in K\}$. Using the convexity and compactness of the set $K$, it is easy to check that the function $f$ is continuous. 
Let $\alpha,\beta\in\Bbb T $ be arbitrary angles. Chose vectors $x,y\in\Bbb R^2$ such that 
$\arg x=\alpha$, $\arg y=\beta $, and $\|x\|=\|y\|=a$. Then $\|x\|=|x|f(\alpha)$ and $\|y\|=|y|f(\beta)$. Also for each $n$ we have $\|x+ny\|=|x+ny|f(\alpha_n)$ and $\|y+nx\|=|y+nx|f(\beta_n)$, where $\lim_{n\to\infty}\alpha_n=\alpha$ and $\lim_{n\to\infty}\beta_n=\beta$. By Theorem of cosines,
$$|x+ny|=|y+nx|=a\sqrt{n^2+1-2n\cos(\alpha-\beta)}=g(n).$$
Since
$$0=\lim_{n\to\infty} \|x+ny\|-\|nx+y\|=\lim_{n\to\infty} g(n)(f(\alpha_n)-f(\beta_n))$$ 
and $\lim_{n\to\infty} g(n)=\infty$, we have $\lim_{n\to\infty}f(\alpha_n)-f(\beta_n)=0$. By the continuity of the function $f$, 
$$f(\alpha)=\lim_{n\to\infty}f(\alpha_n)= \lim_{n\to\infty}f(\beta_n)=f(\beta).$$ 
Thus the function $f$ is constant, so the set $K$ is a standard ball, that implies the claim for the plane.
