Find the distributional limit of $f_n = e^{x/n}$.
That means that I have to find the limit of $T_{f_n}$. I proceeded as follows. Let $\phi \in \mathcal D$ be a test function. Then: $$\langle T_{f_n}, \phi \rangle = \int_{\mathbb R} e^{\frac xn}\phi(x)\,\mathrm dx = -n\int_{\mathbb R} e^{x/n} \phi'(x)\,\mathrm dx$$ where I integrated by parts and exploited the fact that $\phi$ has compact support. From the same fact it follows that the integral is finite (since we also have that $e^{x/n}$ is continuous), and therefore $$\langle T_{f_n}, \phi\rangle \to +\infty$$
I don't know how to interpret this result. In the other exercises, the result was $0$, which is the zero distribution. Here, do I have to conclude that the limit does not exist?
If the limit was a non-zero finite number, should I also say that the limit does not exist? Since a non-zero constant wouldn't identify a distribution.
EDIT: In the sentence above lay my misunderstanding. I was confusing distributions and test functions. A constant $f \equiv c \in \mathbb R$ does indeed identify a distribution, namely $$\langle T_f, \phi \rangle = \int_{\mathbb R} c \cdot \phi(x)\,\mathrm dx \in \mathbb R$$ since $f$ is locally integrable on $\mathbb R$.
Hint $\phi$ has compact support and $e^{\frac{x}{n}} \to 1$ uniformly on compact sets.
P.S. Note that your integration by parts approach cannot help: Indeed, since $e^{x/n} \to 1$ on $supp(\phi')$, we have $$\int_{\mathbb R} e^{x/n} \phi'(x) dx \to \int_{\mathbb R} \phi'(x) dx =0$$ with the last equality following from the fact that the antiderivative of $\phi'$ has compact support.
This means that the limit you get is of the type $\infty \cdot 0$.