PDE text Evans defines Harnack's inequality for non-negative harmonic functions as $$\sup_{B_{R}(0)}u\leq c \inf_{B_{R}(0)}u$$ where $c$ is a constant that only depends on the dimension $n$ such that an open ball $B_{R}(0)\subset \mathbb{R}^n$. Using this inequality is not hard to show that if $u$ is harmonic, non-negative for all $x\in\mathbb{R}^d$, then $u=0.$ By letting $M=\inf_{\mathbb{R}^n}u,$ we can define a new function $v=u-M$, which is non-neagtive and harmonic that saitisfies Harnack's inequality. Since $c$ is independent of $R$ we can take $R\rightarrow\infty$ to get that $u=0$.
But I also saw the following explicit form of Harnack's inequality
$$\frac{R^{d-2}(R-|x|)}{(R+|x|)^{d-1}}u(0)\leq u(x) \leq \frac{R^{d-2}(R+|x|)}{(R-|x|)^{d-1}}u(0)$$
and now I am wondering if the same result can be obtained from this form, by taking $R\rightarrow \infty$, is this justified?