Convergence of $\displaystyle\lim_{s\to1^{-}} (-\Delta)^{s} u(x)=\Delta u(x)$

Question

For $s \in (0,1)$, the fractional Laplacian $(-\Delta)^s u(x)$ can be defined by the singular integral $$ (-\Delta)^s u(x) = c_{n,s} \int_{\mathbb{R}^n} \frac{u(x) - u(y)} {|x-y|^{n+2s}} \mathrm d y , $$ where $c_{n,s}=\frac{4^s\Gamma(n/2+s)}{\pi^{n/2}|\Gamma(-s)|}$ is a constant depending on $n$ and $s$. From this definition, it follows that the $(-\Delta)^s u(x)$ can be characterized in the Fourier space by the Fourier transform $\mathcal{F}$ $$ (-\Delta)^s u(x)=\mathcal{F}^{-1}(2\,\pi|\xi|)^{2s}\, \hat{u}(\xi)). $$ Indeed, we see that $$ (-\Delta) u(x)=\mathcal{F}^{-1}(2\,\pi|\xi|)^{2}\, \hat{u}(\xi)), $$ which is the classical Laplacian acts in a Fourier space as a multiplier of $(2\,\pi|\xi|)^{2}$. See the book "Nonlocal Diffusion and Applications" by Claudia Bucur and Enrico Valdinoci. It immediately follows that $(-\Delta) u(x)$ is the limit case of $(-\Delta)^s u(x)$, i.e. $$ \displaystyle\lim_{s\to1^{-}} (-\Delta)^{s} u(x)=-\Delta u(x). $$ My question is: can we further characterize the rate of convergence in a more precise way, e.g., we have $$ \sup_{x\in\mathbb{R}^n}|(-\Delta)^{s} u(x)+\Delta u(x)|= O(|1-s|^k) $$ for some $k>0$ or in other sense of convergence? Any suggestion for ideas or proofs is welcome, thanks!