TITLE: Limit as x tends to infinity of a product of two functions where one is an integral and the other tends to 0
QUESTION [6 upvotes]: Any hints on how to best approach this problem?
$$\lim_{x\to+\infty} \dfrac{1}{x} \int_{1}^{x} \dfrac{t^3}{1+t^3} dt$$
The first point of confusion for me is that $\dfrac{1}{x} \rightarrow 0$ as $ x \rightarrow +\infty$, so by evaluating the limit for $\dfrac{1}{x}$ and the integral separately and multiplying their limits afterwards should result in $0$, but I highly suspect that this is too simple of a solution that it must be wrong.
Secondly, my hunch is that to evaluate the limit of the integral I could find a function with a smaller area than $\dfrac{t^3}{1+t^3}$ on the listed interval and show that the limit tends to $\infty$ and this would be sufficient to show that $\dfrac{t^3}{1+t^3}$ must also tend to $\infty$ since it has a larger area. Is this the right approach and any hints as to how I could find a function with smaller area that I can show tends to $\infty$?

REPLY [16 votes]: $$\frac{1}{x}\int_{1}^{x}\frac{t^3}{1+t^3}\,dt = \frac{x-1}{x}-\frac{1}{x}\int_{1}^{x}\frac{dt}{1+t^3} = \color{red}{1}+O\left(\frac{1}{x}\right)\quad \text{as }x\to +\infty$$
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.

REPLY [2 votes]: 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...