$\lim_{n\to\infty} f(t/n)=0$ for every $t$ implies $\lim_{t\to0^+}f(t)=0$ (???)

This is a standard application of Baire Category Theorem. Considering $f(\frac 1 t)$ we can state this as follows: f continuous on $(0,\infty)$ and $f(nt) \to 0$ as $n \to \infty$ for all t implies $f(t) \to 0$ as $t \to \infty$. Write $(0,\infty)$ as union of the sets $\{t:|f(nt)| \leq \epsilon$ for all $n \geq m\}$ where $\epsilon >0$ is fixed. Since $(0,\infty)$ is of second category there exists m and $a<b$ such that $(a,b)$ is contained in $\{t:|f(nt)| \leq \epsilon$ for all $n \geq m\}$. Let $x>bm$ and $x(\frac 1a -\frac 1 b)>1$ The interval $(\frac x b ,\frac xa)$ has length exceeding 1 so it contains an integer k. Since $k>\frac x b >m$ and $\frac x k \in (a,b)$ it follows that $|f(k \frac x k)| \leq \epsilon$. Thus $|f(x)| \leq \epsilon$ for $x>bm$ and $x(\frac 1a -\frac 1 b)>1$.