Let $f:\mathbb{R}_+\to \mathbb{R}_+$ be a positive valued continuous function.
I can show that for $y>x>0$ large enough, I have the estimate $$ \int_x^yf(t)dt<1/x.$$
This shows that $f$ is in $\mathbb{L}^1(\mathbb{R}_+)$ and that $\int_x^{+\infty}(t)dt$ decays faster than $1/x$ as $x$ tends to $+\infty.$
My question is: can this be used to show that $f(x)\to 0$ as $x\to+\infty.$
A simple variation of the first answer to the first thread that Martin R linked to gives a counter-example to your desired result: $f(t)$ rises linearly from $0$ to $1$ for $t - \lfloor t \rfloor$ in the interval $\left[0,\frac 1{\lceil t\rceil^2}\right]$, then falls back to $0$ on $\left[\frac 1{2\lceil t\rceil^2},\frac 1{\lceil t\rceil^2}\right]$ then remains at $0$ out to $1$, where the next triangular rise starts.
In each each interval $[n,n+1]$, the area under the function is a triangle of area $\frac 1{4(n+1)^2}$ whose sum from $N$ on is $< \frac 1{2N}$. The function continuous, but the limit as $t \to \infty$ of $f(t)$ does not converge.
It is also evident how to modify the example to show that no similar stronger upper bound on the integral can ever guarantee that $f(t) \to 0$ as $t \to 0$.
