Limit as x tends to infinity of a product of two functions where one is an integral and the other tends to 0

$$\frac{1}{x}\int_{1}^{x}\frac{t^3}{1+t^3}\,dt = \frac{x-1}{x}-\frac{1}{x}\int_{1}^{x}\frac{dt}{1+t^3} = \color{red}{1}+O\left(\frac{1}{x}\right)\quad \text{as }x\to +\infty$$ since $f(t)=\frac{1}{1+t^3}$ is a positive function in $L^1(\mathbb{R}^+)$.
There is no need for de l'Hopital rule or subtler things, simple inequalities do the job just fine.

Note that since $$ \frac{t^3}{1+t^3} > \frac{1}{2} \quad \text{for } t > 1, $$ we clearly have $$ \int_{1}^{x} \dfrac{t^3}{1+t^3} dt \to \infty \quad \text{as } x \to \infty, $$ So you have $$ \lim_{x \to \infty} \frac{\int_{1}^{x} \frac{t^3}{1+t^3} dt}{x} $$ where top and bottom are both infinite and L'Hospital's Rule applies. Then use the Fundamental Theorem of Calculus for the top...