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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 $\infty$?

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    $\begingroup$ $\frac{t^3}{t^3+1}$ tends to $1$, which means that the integral of it tends to $\infty$ as the upper bound grows. $\endgroup$
    – Arthur
    Jul 13, 2017 at 15:52
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    $\begingroup$ Hint: L'Hospital's rule $\endgroup$
    – user281392
    Jul 13, 2017 at 15:53
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    $\begingroup$ 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 $\endgroup$
    – Kelenner
    Jul 13, 2017 at 15:55
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    $\begingroup$ 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, $$\lim_{x\to\infty}\frac{x+\mathcal O(x^0)}x=1$$ $\endgroup$ Jul 13, 2017 at 15:56

2 Answers 2

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

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    $\begingroup$ This is perhaps the simplest approach to the problem. +1 $\endgroup$
    – Paramanand Singh
    Jul 13, 2017 at 16:09
  • $\begingroup$ Very nice (+1). $\endgroup$ Jul 13, 2017 at 16:23
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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...

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    $\begingroup$ 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. $\endgroup$
    – Paramanand Singh
    Jul 13, 2017 at 16:08
  • $\begingroup$ @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. $\endgroup$
    – gt6989b
    Jul 13, 2017 at 16:10
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    $\begingroup$ 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. $\endgroup$
    – Paramanand Singh
    Jul 13, 2017 at 16:24
  • $\begingroup$ @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. $\endgroup$
    – Mark Viola
    Jul 13, 2017 at 17:11

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