Let's say that I want to calculate the Fourier transform of $\frac{1}{|x|^2}$ in $\mathbb{R}^3$. This should not be defined as a function, because $\frac{1}{|x|^2} \notin L^1$, but it should define a tempered distribution. So, knowing that $\mathcal{F}$ is an isomorphism of $\mathcal{S}'$ I consider the functions $e^{- \epsilon |x|} \frac{1}{|x|^2}$, calculate their Fourier transform and take the limit for $\epsilon \to 0$. Turns out that the result is $\frac{C}{|x|}$ for some constant $C$, but that's not important.
What I am asking is if I have understood what I am doing. I have a non-integrable function that, however, can be viewed as a tempered distribution. I consider approximants such as $e^{- \epsilon |x|}\frac{1}{|x|^2}$ that should be viewed as tempered distributions as well. Now, I calculate their Fourier transforms (which makes sense because they are integrable) and see if I have some convergence for $\epsilon \to 0$. Whatever this limit is, I define it to be the Fourier transform of $\frac{1}{|x|^2}$.
Two questions regardng this argument: how can I prove that the distribution to what my approximansts are converging doesn't depend on the particular approximation? And in general, if I had to deal with a generic operator from $\mathcal{S}'$ to $\mathcal{S}'$, this argument should continue to work only if $A$ is continuous. That is, I have a convergent sequence in $\mathcal{S}'$ then its image through $A$ must still converge. Is it right?