Function $f(t)=\frac{t}{2\epsilon}\operatorname{rect}\left(\frac{t}{2\epsilon}\right)$ at $t=0$ My professor said that the function (rect is the rectangular function): $$f(t)=\frac{t}{2\epsilon}\operatorname{rect}\left(\frac{t}{2\epsilon}\right)$$ at $t=0$ is always $0$ for each value of $\epsilon$. Why? If $\epsilon=0$, it will be: $$f(0)=\frac00\operatorname{rect}(0/0)$$ which is an indeterminate form (maybe $\operatorname{rect}(0/0)=\infty$). Thank you for your help.

I will add a subscript $\varepsilon$ to your function $f$.

For any test function $\varphi$, you get $$ \langle \varphi,f_\varepsilon\rangle = \int_{-\varepsilon}^\varepsilon\frac{t}{2\varepsilon}\varphi(t)\,dt, $$ and taking absolute values, $$ \lvert\langle \varphi,f_\varepsilon\rangle\rvert \le\int_{-\varepsilon}^\varepsilon\frac12\lvert\varphi(t)\rvert\,dt \to0\quad\text{as $\varepsilon\to0$}, $$ so that $f_\varepsilon\to0$ in the sense of distributions as $\varepsilon\to0$.

Of course, distributions don't have pointwise values, so it may seem senseless to talk about the value at $t=0$ here, but one can make an exception for distributions arising from continuous functions – in this case, the constant zero.