Convergence of complex exponential in dual Schwartz space I am trying to solve the following problem, but I am a little bit missed. 

Show that the functions $e^{inx}$ and $e^{-inx}$ converge to zero in S' (dual Schwartz space) as $n → ∞$. Conclude that the multiplication of distributions is not a continuous operation, even when it is defined. What is the limit of $\sqrt{n}(1 +n|x|^2)^{-1}$ in $S'$ when $n\rightarrow\infty$?

I would appreciate any help :). 
 A: Convergence in $\mathcal{S}'$ means pointwise convergence. Let $\varphi \in \mathcal{S}$ be a Schwartz function. Now
\begin{align}
\langle e^{inx}, \varphi \rangle &= \int e^{inx} \varphi (x) \, dx = \widehat \varphi(-n)\,.
\end{align}
This tends to $0$ in the limit $n \to \infty$ by the Rieamann–Lebesgue lemma.
Thus $e^{inx} \to 0$ in $\mathcal{S}'$. Similarly you can show that $e^{-inx} \to 0$.
But the product distribution $e^{inx} \cdot e^{-inx}$ should be equal to $1$. Now is $M: \mathcal{S'} \times \mathcal{S'} \to \mathcal{S}'$ is the product operator, meaning
$$
M(f,g) = f \cdot g\,,
$$
where $f$ and $g$ are distributions and $fg$ is the product distribution. Now we have
$$
\lim_{n \to \infty} M(e^{inx}, e^{-inx}) = \lim_{n \to \infty} 1 = 1\,,
$$
and
$$
M( \lim_{n \to \infty} e^{inx}, \lim_{n \to \infty} e^{-inx}) = M(0,0) = 0\,.
$$
Thus $M$ can't be continuous.
You can compute the limit of $\sqrt{n}(1+n|x|^2)^{-1}$ in $\mathcal{S}'$ by considering the limit
$$
\lim_{n \to \infty} \int \sqrt{n} \frac{\varphi(x)}{1+n|x|^2} \, dx\,,
$$
where $\varphi$ is a Schwartz function.
