How to find the following limits in $D'(R)$: $\lim\limits_{k\rightarrow \infty} \cos{kx} P_{\frac{1}{x}}$ and $\lim\limits_{a\rightarrow 0+} e^{ax} P_{\frac{1}{x}}$ ($k\in N$, $a\in R$).
I tried solving this by putting $t=kx$ in first integral and $t=ax$ in second integral, but I'm kind of stuck now. Any kind of help is welcome.
For the second one, note that the multiplication $$ \mathcal{E}\times\mathcal{D}'\longrightarrow \mathcal{D}' $$ is componentwise continuous. In your case this can be explicitly seen, as $$ \lim_{a\to 0} e^{ax}\phi(x) = \phi(x) $$ uniformly in $x$ from the compact set supp $\phi$. The same is true for all derivatives. Hence $$ \lim_{a\to 0} e^{ax} P_{\frac{1}{x}} = P_{\frac{1}{x}}. $$
For the first question using the substitution $kx\mapsto y$ and $\lim_{\epsilon\to 0}\int_{|x|>\epsilon} \frac{f(x)}{x} dx = \int_0^\infty \frac{f(x) - f(-x)}{x} dx$ the integral becomes $$ \int_0^\infty \frac{\cos y}{y} (\phi(\frac{y}{k}) - \phi(-\frac{y}{k})) dy. $$ Splitting this integral into $\int_0^1 ...$ and $\int_1^\infty$, the latter converges to $0$ for $t\to\infty$ by the dominated convergence theorem. Applying the mean value theorem, the first integral becomes $$ \int_0^1\frac{\cos y}{y}\frac{2y}{k} \phi'(\xi_k) dy $$ for some $\xi_k\in [-k,k]$. As $\phi'$ is bounded, for $k\to\infty$ this integral converges to $0$ too. Alltogether, $$ \lim_{k\to\infty} \cos(kx) P_{\frac{1}{x}} = 0. $$