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

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 $\infty$?

• $\frac{t^3}{t^3+1}$ tends to $1$, which means that the integral of it tends to $\infty$ as the upper bound grows. Jul 13, 2017 at 15:52
• Hint: L'Hospital's rule Jul 13, 2017 at 15:53
• Write $t^3/(1+t^3)=1-1/(1+t^3)$ and use that the integral $\int_1^{+\infty}\frac{dt}{1+t^3}$ is convergent Jul 13, 2017 at 15:55
• One approach: $$\int_1^x\frac{t^3}{1+t^3}~\mathrm dt=\int_1^x\frac1{1+\frac1{t^3}}~\mathrm dt=\int_1^x1+\mathcal O(t^{-3})~\mathrm dt=x+\mathcal O(x^0)$$ And, $$\lim_{x\to\infty}\frac{x+\mathcal O(x^0)}x=1$$ Jul 13, 2017 at 15:56

$$\frac{1}{x}\int_{1}^{x}\frac{t^3}{1+t^3}\,dt = \frac{x-1}{x}-\frac{1}{x}\int_{1}^{x}\frac{dt}{1+t^3} = \color{red}{1}+O\left(\frac{1}{x}\right)\quad \text{as }x\to +\infty$$ since $f(t)=\frac{1}{1+t^3}$ is a positive function in $L^1(\mathbb{R}^+)$.
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...
• Your don't need to ensure that the integral tends to $\infty$ in order to use L'Hospital's Rule. Just denominator $x$ tending to $\infty$ is sufficient. Jul 13, 2017 at 16:08
• @ParamanandSingh I don't think that's true. L'Hospital's Rule applies to indeterminate forms, either $0/0$ or $\infty/\infty$. If the integral converges, for example, the rule does not apply. Jul 13, 2017 at 16:10
• Like many you are not aware that the rule applies for $0/0$ and "$\text{anything} /\infty$" scenarios. Check Wikipedia article en.wikipedia.org/wiki/L%27H%C3%B4pital%27H rule and try to the proof given there. Jul 13, 2017 at 16:24
• @gt6989b Paramanand is correct. LHR applies to $\frac{\text{anything}}{\infty}$, even if the limit of the "anything" fails to exist, provided the other conditions of LHR hold. Jul 13, 2017 at 17:11